Projectile Motion Notes. Vertical Projectile Motion.

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Projectile Motion Notes
Vertical Projectile Motion
Vertical Projectile Motion
Case 1 Case 2 Case 3 Case 4
u
u
uObject dropped from rest
Object projected downwards
Object launched vertically and
returns to launch height
Object launched vertically and returns to a
different height
Case 1: Vertical Projectile Motion
+
Three points
Set positive direction as down
If time not involved can use energy approachUg Ek
mgh = ½mv2
gh = ½v2
2gh = v2
Use constant acceleration equations with
u = 0 a = 10ms1
Case 2: Vertical Projectile Motion
+
Three points
Set positive direction as down
Use constant acceleration equations with
a = 10ms1u
If time not involved can use energy approachEk
i + Ug Ek
f
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
Case 3: Vertical Projectile Motion
Six points
+Set positive
direction as up u
Speeds up are the same as down at
each height
v
v
v = 0
If time not involved can use energy approach
Eki Ek
f + Ug
½mu2 = ½mv2 + mgh½u2 = ½v2 + ghu2 = v2 + 2gh
Use const accel equations with a = 10ms1
Time up = time down or
Total time = 2 × time up
Case 4: Vertical Projectile Motion
Six points
+Set positive
direction as up
v = 0
Use const accel equations with a = 10ms1
Speeds up are the same as down at
each height
If time not involved can use energy approach
Eki Ek
f + Ug
½mu2 = ½mv2 + mgh
Displacements below launch height will be
negative
u
Case 4: Vertical Projectile Motion
Six points
+Set positive
direction as up
v = 0
Use const accel equations with a = 10ms1
Speeds up are the same as down at
each height
If time not involved can use energy approach
Eki Ek
f + Ug
½mu2 = ½mv2 + mgh
Displacements below launch height will be
negative
u
Vertical Projectile Motion
Worked Examples
Vertical Projectile Motion
Example 1A stone is dropped from an 8.0m tower and falls to the ground.(a) How long will it take the stone to drop to the ground?
t = ? u = 0x = 8.0m a = 10ms2
x = ut + ½ at2
8 = ½ × 10 × t2
8 = 5 × t2
1.6 = t2
1.26491 = t
t 1.3 s
+
• In Maths when you determine the square root of a number you should always give the result as a value. This means there are two answers
• In Physics we normally just write out the magnitude of the square root because the negative value is either not relevant or it represents a direction which is already obvious in the problem. In this case a negative time is not plausible.
Vertical Projectile Motion
Example 1A stone is dropped from an 8.0m tower and falls to the ground.(b) What speed will the stone hit the ground?
v = ? u = 0x = 8.0m a = 10ms2
v2 = u2 + 2ax
v2 = 2 × 10 × 8
v2 = 160
v = 12.6491
v 13 ms1
+
Alternative – Energy Approach
Ug Ek
mgh = ½mv2
gh = ½v2
2gh = v2
2 × 10 × 8 = v2
160 = v2
12.6491 = vv = 13ms1
• In In Physics we normally just write out the magnitude of the square root because the negative value is either not relevant or it represents a direction which is already obvious in the problem. In this case a negative value means the stone is moving upwards which is not plausible.
Vertical Projectile Motion
Example 2A stone is dropped down a well and takes 2.5 seconds to hit the water. How deep is the well?
x = ? u = 0 t = 2.5s a = 10ms2
x = ut + ½ at2
x = ½ × 10 × 2.52
x = 31.25
x 31 s+
Vertical Projectile Motion
Example 3A stone is thrown down at 2.0ms1 from a 3.0m tower and falls to the ground. What speed did the ground hit the ground?
v = ? u = 0x = 10m a = 10ms2
v2 = u2 + 2ax
v2 = 22 + 2 × 10 × 3
v2 = 64
v = 8 ms1
+ Alternative – Energy Approach
Eki + Ug Ek
f
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
22 + 2 × 10 × 3 = v2
64 = v2
8 = vv = 8ms1
2.0ms1
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(a) How high will the cannon ball reach?
x = ? u = 30ms1 v = 0 a = –10ms2
v2 = u2 + 2ax
0 = 302 + 2 × – 10 × x
0 = 900 + – 20 × x– 900 = – 20 × x
45 = x
x = 45m
+30ms1
x
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(a) How high will the cannon ball reach?
x = ? u = 30ms1 v = 0 a = –10ms2
v2 = u2 + 2ax
0 = 302 + 2 × – 10 × x
0 = 900 + – 20 × x– 900 = – 20 × x
45 = x
x = 45m
+30ms1
x
Alternative – Energy Approach
EkUg ½mv2 = mgh
½v2 = ghv2 = 2gh
302 = 2 × 10 × h 900 = 20h
45 = hh = 45m
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(b) What will be the flight time of the cannon ball?
Find time to top of the path
t = ? u = 30ms1 v = 0 a = –10ms2
v = u + at
0 = 30 + – 10 × t– 30 = – 10 × t
3 = t
Find total time
Total time = 2 × 3
= 6 seconds
+30ms1
t t
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(b) What will be the flight time of the cannon ball?
Find time to top of the path
t = ? u = 30ms1 v = 0 a = –10ms1
v = u + at
0 = 30 + – 10 × t– 30 = – 10 × t
3 = t
Find total time
Total time = 2 × 3
= 6 seconds
+30ms1
t t
Alternative 1t = ? u = 30ms1 x = 0 a = – 10ms2
x = ut + ½at2
0 = 30t + ½ × – 10 × t2
0 = 30t + – 5 × t2
0 = 5t (6 – t)t = 0, 6So flight time = 6 seconds
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(b) What will be the flight time of the cannon ball?
Find time to top of the path
t = ? u = 30ms1 v = 0 a = –10ms1
v = u + at
0 = 30 + – 10 × t– 30 = – 10 × t
3 = t
Find total time
Total time = 2 × 3
= 6 seconds
+30ms1
t t
Alternative 2t = ? u = 30ms1 v = – 30ms1 a = – 10ms2
v = u + at– 30 = 30 + – 10 × t– 60 = – 10 × t 6 = t
So flight time = 6 seconds
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(c) What will be the speed at a height of 30m?
t = ? u = 30ms1 x = 30m a = –
10ms2
v2 = u2 + 2ax
v2 = 302 + 2 × – 10 × 30
v2 = 300
v = 17.321 ms1
So the speed will be 17ms1 at 30m on the
way up and 17ms1 at 30m the way down
+30ms1
30m
v
v
Vertical Projectile Motion
Example 4A cannon ball is fired upwards from the ground at a speed of 30ms1.(c) What will be the speed at a height of 30m?
t = ? u = 30ms1 x = 30m a = –
10ms2
v2 = u2 + 2ax
v2 = 302 + 2 × – 10 × 30
v2 = 300
v = 17.321 ms1
So the speed will be 17ms1 at 30m on the
way up and 17ms1 at 30m the way down
+30ms1
30m
v
v
Alternative Using energy approach
v = ? u = 30ms1 h = 30m g = 10ms1
Eki Ek
f + Ug ½mu2 = ½mv2 + mgh
½u2 = ½v2+ gh u2 = v2+ 2gh
302 = v2+ 2×10×30 900 = v2 + 600
300 = v2
17.3205 = vv 17 ms1
Vertical Projectile Motion
Example 5If a cannon ball fired from 4.0m above the ground is in the air for 8.0 seconds, what is the launch speed of the cannon ball?
u = ? x = – 4.0m t = 8.0s a = –10ms2
x = ut + ½ at2
– 4 = u × 8 + ½ × – 10 × 82
– 4 = 8u – 320
316 = 8u
39.5 = u
u 40 ms1
+
4.0m
Vertical Projectile Motion
Exam Questions
2006 Exam
Q6 & 7A rocket of mass 0.50 kg is set on the ground, pointing vertically up. When ignited, the gunpowder burns for a period of 1.5 s, and provides a constant force of 22 N. The mass of the gunpowder is very small compared to the mass of the rocket, and can be ignored. After 1.5 s, what is the height of the rocket above the ground?
+22N
W = mg = 0.5× 10 = 5N
v = ? t = 1.5s u = 0 a = = = 34ms2 v = u + atv = 0 + 34 × 1.5v = 51 ms1
alternative using momentum/impulse approachv = ? t = 1.5s u = 0
Fnet = 17NFt = m(v – u)17 × 1.5 = 0.5v25.5 = 0.5v51 ms1 = v
When you have F, time and no distance you can quite often use formulae from the momentum/impulse string
I = Ft = p = pf – pi = m(v– u)to solve the problem
Horizontally Launched Projectile Motion
Horizontally Launched Projectile Motion
Case 1 Case 2
Horizontally launched object drops to ground levelEx1 – object launched of
building/cliffEx2 – object falling off a
moving vehicleEx3 – object dropped from
an aircraft
uuu
Horizontally Launched Projectile Motion
Case 1 Case 2
uuHorizontally launched object drops to another heightEx – object launched of
building to the top of another building
Case 2: Horizontally Launched Projectile Motion
u
Six points
Reference axes
y
x
In y direction uy = 0 a = 10ms1
and use const accel equns
If time not involvedEk
i + Ug Ek
f
½mu2 + mgh = ½mv2
u2 + 2gh = v2
In x dirn vx = u and use
dv t
If not enough information in one direction get time from the
other direction
vx
vyvx
vy
vx
vy
The motion in the x & y directions are independent
vy is accelerated by gvx is constant = u
Case 2: Horizontally Launched Projectile Motion
u
Six points
y
x
Four landing formulae
On landing
thproj=
Rhproj = u
v =
= tan1
vx
vyvx
vy
vx
vy
What do and u represent?
vx & vy
on landing
v
u
Case 2: Horizontally Launched Projectile Motion
Six points
If you treat the top of the second building as a virtual ground level all the previous five points including formulae hold for this situation
h1h2
h = h1 – h2
Virtual ground level
Four landing formulae
Horizontal Projectile Motion
Worked Examples
Horizontally Launched Projectile Motion
Example 1An aeroplane travelling at 360kmh1 drops a food package when it passes over a hiker at a height of 300m.
(a) How far from the hiker will the food land?
In x direction
d = ? v = ux = 100ms1 t = ?
d = v t
d = 100 × 7.74597
d = 774.597
d 775m
Find t from the y directiont = ? u = 0 x=8.0m a = 10ms1
x = ut + ½ at2
300= ½ × 10 × t2
300 = 5 × t2
60 = t2
7.74597 = t
7.74597s
360kmh1
= 100ms1
y
x
d = ?
Horizontally Launched Projectile Motion
Example 1An aeroplane travelling at 360kmh1 dropped a food package when it passes over a hiker at a height of 300m.
(a) How far from the hiker will the food land?
In x direction
d = ? v = ux = 100ms1 t = ?
d = v t
d = 100 × 7.74597
d = 774.597
d 775m
Find t from the y directiont = ? u = 0 x=8.0m a = 10ms1
x = ut + ½ at2
300= ½ × 10 × t2
300 = 5 × t2
60 = t2
7.74597 = t
7.74597s
360kmh1
= 100ms1
y
x
Alternative using derived formulaeRhproj = ? u = 100ms1 h = 300m g = 10ms1
Rhproj = uRhproj = Rhproj = 774.597Rhproj 775m
d = ?
Horizontally Launched Projectile Motion
Example 1An aeroplane travelling at 360kmh1 dropped a food package when it passes over a hiker at a height of 300m.
(b) What will be the velocity of the package when it hits the ground?
In y directionvy = ? uy = 0 x=300m a = 10ms1
v2 = u2 + 2ax
v2 = 2 × 10 × 300
v2 =6000
v = 77.4597
Find final velocity
h2 = a2 + b2 v2 = 1002 + 77.45072
v2 = 16000v = 126.49v 126 ms1
360kmh1
= 100ms1
y
x
100ms1
77.4597ms1
tan = tan = = tan1 = 37.761o
38o
v
38o
126 ms1
v = ?
Horizontally Launched Projectile Motion
Example 1An aeroplane travelling at 360kmh1 dropped a food package when it passes over a hiker at a height of 300m.
(b) What will be the velocity of the package when it hits the ground?
In y direction
vy = ? uy = 0 x=300m a = 10ms1
v2 = u2 + 2ax
v2 = 2 × 10 × 300
v2 =6000
v = 77.4597
Find final velocity
h2 = a2 + b2 v2 = 1002 + 77.45072
v2 = 16000v = 126.49v 126 ms1
360kmh1
= 100ms1
y
x
100ms1
77.4597ms1
tan = tan = = tan1 = 37.761o
38o
v38o
126 ms1Alternative using derived formulaev = ? u = 100ms1 h = 300m g = 10ms1
v = v = v = 126.491v 126ms1
= tan1 = tan1 = 37.761o 38o
38o
126 ms1
v = ?
Horizontally Launched Projectile Motion
Example 1An aeroplane travelling at 360kmh1 dropped a food package when it passes over a hiker at a height of 300m.
(b) What will be the velocity of the package when it hits the ground?
In y direction
vy = ? uy = 0 x=300m a = 10ms1
v2 = u2 + 2ax
v2 = 2 × 10 × 300
v2 =6000
v = 77.4597
Find final velocity
h2 = a2 + b2 v2 = 1002 + 77.45072
v2 = 16000v = 126.49v 126 ms1
360kmh1
= 100ms1
y
x
100ms1
tan = tan = = tan1 = 37.761o
38o
v38o
126 ms1
v = ?
Alternative if just asked for speed (energy approach)v = ? u = 100ms1 h = 300m g = 10ms1
Eki + Ug Ek
f
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
1002 + 2×10×300 = v2
16000 = v2
126.491 = vv 126ms1
Horizontally Launched Projectile Motion
Example 1(c) The aeroplane has to do another drop
but from a height 180m. What is the maximum speed (in kmh1) that the plane can have for the food to land within 300m from the hiker if the parcel is again released when the plane is over the hiker?
In x direction
v = ux = ? d= 300m t = ?
v =
v =
v = 50ms1
v = 180kmh1
Find t from the y directiont = ? u = 0 x= 180m a = 10ms1
x = ut + ½ at2
180= ½ × 10 × t2
180 = 5 × t2
36 = t2
6 = t
6s
v = kmh1
y
x
d = 300m
Horizontally Launched Projectile Motion
Example 1(c) The aeroplane has to do another drop
but from a height 180m. What is the maximum speed (in kmh1) that the plane can have for the food to land within 300m from the hiker if the parcel is again released when the plane is over the hiker?
In x direction
v = ux = ? d= 300m t = ?
v =
v =
v = 50ms1
v = 180kmh1
Find t from the y directiont = ? u = 0 x= 180m a = 10ms1
x = ut + ½ at2
180= ½ × 10 × t2
180 = 5 × t2
36 = t2
6 = t
6s
v = kmh1
y
x
d = 300m
Alternative using derived formulaeu = ? Rhproj = 300 h = 180ms1 g = 10ms2
Rhproj = u300 = 300 = × 6 50 = uu = 180kmh1
Horizontally Launched Projectile Motion
Example 2A car is driven off roof of a 15m building at 10ms1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?
Find h2 from x direction
x =? u = 0 a = 10ms2 t = ?
x = ut + ½ at2
x = ½ × 10 × 0.52
x = 1.25mFind height of second building
Height = 15 – 1.25
= 13.75
14m
Find t from the x directiont = ? v = ux = 10ms1 d= 5.0m
t = t = t = 0.5 s
0.5s
y
x20ms1
15mh1
h2
5.0m
Not drawn to scale
Horizontally Launched Projectile Motion
Example 2A car is driven off roof of a 15m building at 10ms1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?
Find h2 from x direction
x =? u = 0 a = 10ms2 t = ?
x = ut + ½ at2
x = ½ × 10 × 0.52
x = 1.25mFind height of second building
Height = 15 – 1.25
= 13.75
14m
Find t from the x directiont = ? v = ux = 10ms1 d= 5.0m
t = t = t = 0.5 s
0.5s
y
x10ms1
15mh1
h2
5.0m
Not drawn to scaleAlternative using derived formulaeh = h2 = ? Rhproj = 5.0 u = 10ms1 g = 10ms1
Rhproj = u5 = 0.5 =
0.25 = 1.25 = h
Find height of second buildingHeight = 15 – 1.25 = 13.75 14m
Horizontal Projectile Motion
Exam Questions
2002 Exam Q5
Q1A sportscar is driven horizontally off building 1 and lands on a building 20m away. The floor where the car lands in building 2 is 4.0 m below the floor from which it started in building 1. Calculate the minimum speed at which the car should leave building 1 in order to land in the car park of building .
In x directionv = ux = ? d= 20m t = ?v = v = v = 22.3607ms1
v 22 ms1
Find t from the y directiont = ? u = 0 x= 4m a = 10ms2
x = ut + ½ at2
4= ½ × 10 × t2
4 = 5 × t2
0.8 = t2
0.89443 = t
0.89443s
2002 Exam Q5
Q1A sportscar is driven horizontally off building 1 and lands on a building 20m away. The floor where the car lands in building 2 is 4.0 m below the floor from which it started in building 1. Calculate the minimum speed at which the car should leave building 1 in order to land in the car park of building .
In x directionv = ux = ? d= 20m t = ?v = v = v = 22.3607ms1
v 22 ms1
Find t from the y directiont = ? u = 0 x= 4m a = 10ms2
x = ut + ½ at2
4= ½ × 10 × t2
4 = 5 × t2
0.8 = t2
0.89443 = t
0.89443s
Alternative using derived formulaeu = ? Rhproj = 20.0 h = 4.0ms1 g = 10ms2
Rhproj = u20 = 20 = × 0.89443
22.3607 = uu 22 ms1
2002 Exam Q6
Q2In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of 25 ms–1
Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.
In y directionvy = ? uy = 0 x=4.0m a = 10ms2
v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427
Find final velocity
h2 = a2 + b2 v2 = 252 + 8.9944272
v2 = 705v = 26.5518v 27 ms1
25ms1
8.94427ms1
v
2002 Exam Q6
Q2In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of 25 ms–1
Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.
In y directionvy = ? uy = 0 x=4.0m a = 10ms2
v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427
Find final velocity
h2 = a2 + b2 v2 = 252 + 8.9944272
v2 = 705v = 26.5518v 27 ms1
25ms1
8.94427ms1
tan = tan = = tan1 = 18.96173o
19o
v
19o
27 ms1
tan = tan = = tan1 = 37.761o
38o
v38o
126 ms1Alternative using Derived Formulav = ? u = 25ms1 h = 4.0m g = 10ms1
v = v = v = 26.5518v 27ms1
2002 Exam Q6
Q2In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of 25 ms–1
Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.
In y directionvy = ? uy = 0 x=4.0m a = 10ms2
v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427
Find final velocity
h2 = a2 + b2 v2 = 252 + 8.9944272
v2 = 705v = 26.5518v 27 ms1
25ms1
8.94427ms1
tan = tan = = tan1 = 18.96173o
19o
v
19o
27 ms1
tan = tan = = tan1 = 37.761o
38o
v38o
126 ms1Alternative using energy approachv = ? u = 25ms1 h = 4.0m g = 10Nkg1
Eki + Ug Ek
f
½mu2 + mgh = ½mv2
u2 + 2gh = v2
v =
v = v = 26.5518v 27ms1
Same as derived formula
2006 Exam Q8
Q3A 0.5kg rocket is launched horizontally and propelled by a constant force of 22 N for 1.5 s, from the top of a 50 m tall building. Assume that in its subsequent motion the rocket always points horizontally. After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?
22N
y
x
In x directionv = ? t = 1.5s u = 0 a = = = 44ms2 v = u + atv = 0 + 44 × 1.5v = 66 ms1
In y directionv = ? t = 1.5s u = 0 a = 10ms2 v = u + atv = 0 + 10 × 1.5v = 15 ms1
alternative using momentum/impulse approachv = ? t = 1.5s u = 0
Fnet = 22NFt = m(v – u)22 × 1.5 = 0.5v33 = 0.5v66 ms1 = v
2006 Exam Q8
Q3After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?
22N
y
x
Find final velocity
h2 = a2 + b2 v2 = 662 + 152
v2 = 4581v = 67.6831v 68 ms1
66ms1
15ms1
tan = tan = = tan1 = 12.8043o
13o
v
13o
68 ms1
Obliquely Launched Projectile Motion
Obliquely Launched Projectile Motion
Case 1: landing height = launch heightCase 2: landing height different to launch height
Case 1: Obliquely Launched Projectiles Landing At The Launching Height
y
x
u
uy = u sin
ux = u cos
Reference axes
In y direction uy = u sin a = –10ms1
use const accel equns
In x dirn vx = u cos in
If not enough info in one dirn get time from the other dirn
The motion in the x & y directions are independent
uy = u sin vx is constant = u cos
dv t
vx
vy
vx
vy
vx
Eight points
If time not involvedEk
i Ek
f + Ug ½mu2 = ½mv2+ mgh
u2 = v2+ 2gh
vyv = u=
vx
Speeds before hmax are the same as those after
at each height
Ekmin = ½ mvx
2
Case 1: Obliquely Launched Projectiles Landing At The Launching Height
y
x
u
uy = u sin
ux = u cos
vx
vy
vx
vx
vy
vx
vyv = u=
Three formulae
tflight =range =hmax=
Case 2: Obliquely Launched Projectiles Landing At A Different Height To The Launching Height
Notes for Case 2• The eight points relating to oblique projectiles that land at the
launching height also apply to this type of situation. • You will not be asked to find the time from the vertical axis since it
will involve solving a quadratic equation. Instead you will either be given the time or asked to work it out from a horizontal distance (using t =
ux
Obliquely Launched Projectile Motion
Worked Examples
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(a) How far will the ball travel before it strikes the ground if the ground is flat?
In x direction
d = ? v = 28.096ms1 t = ?
Find t to top of path from y dirnt = ? u = 13.101 v =0 a = –10ms2
v = u + at 0 = 13.101 + – 10 × t
– 13.101 = – 10 × t 1.3101 = t
Find total time Total time = 2 × 1.3101 = 2.6202s
25o
31ms1
y
x
uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
d
What are the three other ways of working out the flight time?1. t = ? u = 13.101 x =0 a = –10ms2 (this will
involve a factorisation)2. t = ? u = 13.101 v = –13.101 a = –10ms2
(this is pretty easy)3. Using flight time derived formula (just put the
numbers in and work out)
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(a) How far will the ball travel before it strikes the ground if the ground is flat?
In x direction
d = ? v = 28.096ms1 t = ?
Find t from y dirnt = ? u = 13.101 x =0 a = –10ms2
x = ut + ½at2
0 = 13.101 t + ½ × – 10 × t2
0 = 13.101 t – – 5 t2
0 = t (13.101 – 5t ) t = 0,
t = 2.6202
y
x uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
What are the three other ways of working out the flight time?1. t = ? u = 13.101 x =0 a = –10ms2 (this will
involve a factorisation)2. t = ? u = 13.101 v = –13.101 a = –10ms2
(this is pretty easy)3. Using flight time derived formula (just put the
numbers in and work out)
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(a) How far will the ball travel before it strikes the ground if the ground is flat?
In x direction
d = ? v = 28.096ms1 t = ?
Find t from y dirnt = ? u = 13.101 v =13.101 a = –10ms2
v = u + at 13.101 = 13.101 + – 10 × t
– 26.202 = – 10 × t 2.6202 = t
y
x uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
What are the three other ways of working out the flight time?1. t = ? u = 13.101 x =0 a = –10ms2 (this will
involve a factorisation)2. t = ? u = 13.101 v = –13.101 a = –10ms2
(this is pretty easy)3. Using flight time derived formula (just put the
numbers in and work out)
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(a) How far will the ball travel before it strikes the ground if the ground is flat?
In x direction
d = ? v = 28.096ms1 t = ?
d = v t
d = 28.096 × 2.6202
d = 73.617
d 74m
Find t from flight formulat = ? u = 31ms1 = 25o g = –10ms2
tflight =tflight =tflight = 2.6202
2.6202s
y
x uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
What are the three other ways of working out the flight time?1. t = ? u = 13.101 x =0 a = –10ms2 (this will
involve a factorisation)2. t = ? u = 13.101 v = –13.101 a = –10ms2
(this is pretty easy)3. Using flight time derived formula (just put the
numbers in and work out)
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(a) How far will the ball travel before it strikes the ground if the ground is flat?
In x direction
d = ? v = 28.096ms1 t = ?
d = v t
d = 28.096 × 2.6202
d = 73.617
d 74m
Find t to top of path from y dirnt = ? u = 13.101 v =0 a = –10ms2
v = u + at 0 = 13.101 + – 10 ×
– 13.101 = – 10 × t 1.3101 = t
Find total time Total time = 2 × 1.3101 = 2.6202s
2.6202s
1 step alternative using derived formulaeRobliq = ? u = 31ms1 = 25o g = 10ms1
Robliq =Robliq =Robliq = 73.617Robliq 74m
25o
31ms1
range
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(b) What is the minimum speed of the golf ball?
Minimum speed occurs at the top of the path when there only the horizontal component of velocity.
So minimum is 28.096ms1
25o
31ms1
y
x
uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(c) How high will the golf ball rise?
In y directionx = ? u = 13.101ms1 v = 0 a = –10ms2
v2 = u2 + 2ax0 = 13.1012 + 2 × – 10 × x0 = 171.636 + – 20 × x
– 171.636 = – 20 × x 8.5818 = x
x = 8.6m
25o
31ms1
y
x
uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(c) How high will the golf ball rise?
In y directionx = ? u = 13.101ms1 v = 0 a = –10ms2
v2 = u2 + 2ax0 = 13.1012 + 2 × – 10 × x0 = 171.636 + – 20 × x
– 171.636 = – 20 × x 8.5818 = x
x = 8.6m
Alternative using derived formulaehmax= ? u = 31ms1 = 25o g = 10ms1
hmax= hmax=
hmax =8.5820hmax 8.6m
25o
31ms1
y
x
uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(d) What is the velocity of the ball as it hits the ground?
Since the ball hits the ground at the same height as it launch height final speed is:
25o
31ms1
25o
31ms1
y
x
uy = 31sin25 = 13.101ms1
ux = 31cos25 = 28.096ms1
Obliquely Launched Projectile Motion
Example 1A golf ball is hit from the ground at 31ms1 at 25o to the horizontal.
(e) What will be the speed of the ball 20m off the ground?
Using energy approach (which avoids having to work with vectors)
v = ? u = 31ms1 h = 20m g = 10ms1
Eki Ek
f + Ug ½mu2 = ½mv2 + mgh
½u2 = ½v2+ gh u2 = v2+ 2gh
312 = v2+ 2×10×20 961 = v2 + 400
561 = v2
23.685 = vv 24 ms1
25o
31ms1
y
x
Obliquely Launched Projectile Motion
Example 2A golf ball is hit at 25ms1 at 35o to the horizontal from a cliff edge and it takes 6.0 seconds to hit the ground below the cliff.
(a) How high is the cliff?
In y direction x = ? u = 14.3390ms1 t = 6.0s a = –10ms2
x = ut + ½ at2
x = 14.3390 × 6 + ½ × – 10 × 62
x = – 93.966
so the cliff is 94m high
35o
25ms1
y
x
uy = 25sin35 = 14.339ms1
ux = 25cos35 = 20.4796ms1
x
Obliquely Launched Projectile Motion
Example 2A golf ball is hit at 25ms1 at 35o to the horizontal from a cliff edge and it takes 6.0 seconds to hit the ground below the cliff.
(b) How far from the base of the cliff does the ball land?
In x direction d = ? v = 20.4796ms1 t = 6
d = v t
d = 20.4796 × 6
d = 122.875
d 123m
35o
25ms1
y
x
uy = 25sin35 = 14.339ms1
ux = 25cos35 = 20.4796ms1
d
Obliquely Launched Projectile Motion
Example 3A golf ball is hit at 28ms1 at 60o to the horizontal from a fairway and hits a tree that is 56m away. How far up the tree does the golf ball strike the tree?
In y direction
x = ? u = 24.249ms1 a = –10ms2 t = ?
x = ut + ½ at2
x = 24.249 × 4 + ½ × – 10 × 42
x = 16.996
so the ball hits the tree 17m up
60o
28ms1
y
x
uy = 28sin60 = 24.249ms1
ux = 28cos60 = 14ms1
56m
h
Find t from the x directiont = ? v = ux = 14ms1 d= 56m
t = t = t = 4 s
4s
28ms1
60o
Obliquely Launched Projectile Motion
Exam Questions
2004 Exam Q7
Q1The diagram shows a motorcycle rider using a 20° ramp to jump her motorcycle across a river that is 10.0 m wide.Question 7Calculate the minimum speed that the motorcycle and rider must leave the top of the first ramp to cross safely to the second ramp that is at the same height. (The motorcycle and rider can be treated as a pointparticle.) u = ? Robliq =10 = 20o g = 10ms2
Robliq =
10 =
100 = u2sin40
155.572 = u2
12.4729 = u
u 12 ms1
Using x & y direction analysis for this problem is very difficult so using the range formula is a much more practical way to solve this problem
2005 Exam Q11
Q2
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s1.With what speed, relative to the deck, did the ball leave Fred's racket? Give your answer to three significant figures.
yx
8o
u
30ms1
30ms1
3.0m
8o
30ms1
cos =
cos 8 =
u =
u = 30.29483ms1
u 30.3ms1
2005 Exam Q12
Q3
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s1.At its highest point, how far was the ball above the ground.
x = ? u = 30ms1 v = 0 a = –10ms2
v2 = u2 + 2ax
0 = 4.2162252 + 2 × – 10 × x
0 = 17.77655 + – 20 × x– 17.77655 = – 20 × x
0.888828 = x
Overall height = 0.888828 + 3
= 3.888828
3.89 m
yx
8o
30.29483ms1
30ms1
30ms1
3.0m
8o
30ms1
30.29483sin8= 4.216225ms1
x
3.0m
2005 Exam Q12
Q3
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s1.At its highest point, how far was the ball above the ground.
x = ? u = 30ms1 v = 0 a = –10ms2
v2 = u2 + 2ax
0 = 4.2162252 + 2 × – 10 × x
0 = 17.77655 + – 20 × x– 17.77655 = – 20 × x
0.888828 = x
Overall height = 0.888828 + 3
= 3.888828
3.89 m
yx
8o
30.29483ms1
30ms1
30ms1
3.0m
8o
30ms1
hmax
3.0m
Alternative using derived formulaehmax= ? u = 30.29483ms1 = 8o g = 10ms2
hmax= hmax=
hmax = 0.888828
2005 Exam Q12
Q3
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s1.At its highest point, how far was the ball above the ground.
x = ? u = 30ms1 v = 0 a = –10ms2
v2 = u2 + 2ax
0 = 4.2162252 + 2 × – 10 × x
0 = 17.77655 + – 20 × x– 17.77655 = – 20 × x
0.888828 = x
Overall height = 0.888828 + 3
= 3.888828
3.89 m
yx
8o
30.29483ms1
30ms1
30ms1
3.0m
8o
30ms1
h2
Alternative using Energy Approachh2= ? u = 30.29483ms1 h1 = 3m g = 10ms2
Eki + Ug
i Ekf + Ug
f
½ mu2 + mgh1 = ½ mv2 + mgh2
½ u2 + gh1 = ½ v2 + gh2
½ × 30.294832 + 10 × 3 = ½ × 302 + 10 × h2
488.888 = 450 + 10 × h2
3.8888 = h2
h2 3.89m
h1 =
2007 Exam Q14
Q4Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the time of flight of the paintball?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
Find t from the x directiont = ? v = ux = 28.096ms1 d= 127m
t = t = t = 3.5032 st 3.5 s
2007 Exam Q14
Q4Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the time of flight of the paintball?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
AlternativeFind t from the y directiont = ? u = 16.9047ms1 v = –16.9047ms1 a = –10ms2
v = u + at  16.9047 = 16.9047 + – 10 × t – 33.8094 = – 10 × t 3.38094 = t t 3.4 s
There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction
2007 Exam Q14
Q4Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the time of flight of the paintball?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
Find t from the y directiont = ? u = 16.9047ms1 v = –16.9047ms1 a = –10ms2
v = u + at  16.9047 = 16.9047 + – 10 × t – 33.8094 = – 10 × t 3.38094 = t t 3.4 s
There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction
2007 Exam Q14
Q4Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the time of flight of the paintball?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
Alternative with formulaFind t from flight formula t = ? u = 40ms1 = 25o g = –10ms2
tflight =tflight =tflight = 3.3809tflight 3.4s
There is a discrepancy in the data for this problem hence the slight difference in answers from the previous methods
2007 Exam Q15
Q5Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the value of h, the maximum height above the firing level?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
Find hmax from y direction x = ? u = 16.9047ms1 v = 0 a = –
10ms2 v2 = u2 + 2ax0 = 16.90472 + 2 × – 10 × x0 = 287.201+ – 20 × x
– 287.201 = – 20 × x 14.3600 = x
x 14m
2007 Exam Q15
Q5Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.What is the value of h, the maximum height above the firing level?
25o
40ms1y
x uy = 40sin25 = 16.9047ms1
ux = 40cos25 = 36.2523ms1
Alternative with formulaFind t from flight formula
hmax= ? u = 40ms1 = 25o g = 10ms2
hmax= hmax=
hmax = 14.2885hmax 14m
There is a discrepancy in the data for this problem hence the slight difference in answers from the previous method
2007 Exam Q16
Q6Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was fired are the same.Which of the following diagrams (A–D) below gives the direction of the force acting on the paintball at points X and Y respectively?
Since air resistance is ignored the only force on the paintball is gravity which is constant and downwards,
so the acceleration must be constant and downwards.
2007 Exam Q17
Q7Later in the game, Daniel is twice as far away from John (254 m). John fires an identical paintball from the same height above the ground as before. The ball hits Daniel at the same height as before. In both cases the paintball reaches the same maximum height (h) above the ground.
Which one or more of the following is the same in both cases?A. flight time B. initial speed C. accelerationD. angle of firing level?
Because the paintball gets to the same vertical height the vertical component in each
situation must be the same and hence the flight time will also be the same.
Since air resistance is ignored the only force on the paintball is gravity which is constant in both situations and so the acceleration is constant
as well.
So With More Than One Method For Solving Most Projectile
Motion Questions – How Do I Decide On The Best?
Some Loose Rules To Follow.
Some Loose Rules For Deciding On The Best Method To Solve A Problem
1. The best method is the one that makes the most sense to you and doesn’t take forever to get to an answer
2. Often if there is a formula use a formula, but be careful of expressions such as• hmax== 0.888828
• Robliq =
• v = = 126.491
• Rhproj = = 774.597
3. Make sure you practice more than one method when solving projectile motion questions – particularly working with x & y directions.
Some Loose Rules For Deciding On The Best Method To Solve A Problem
4. An energy approach is very useful when asked for the magnitude of a velocity that is not at launching height or maximum height.
5. Remember there are three approaches that can be used for solving
Constant Accel Equations
Impulse/ Momentum
Formulae(no dist)
Energy Formulae(no time)