Projectile Motion Notes. Vertical Projectile Motion.

Projectile Motion Notes

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 = 10ms-1


+

Three points

Set positive direction as down

Use constant acceleration equations with

a = 10ms-1u

If time not involved can use energy approachEk

i + Ug Ek

f

½mu2 + mgh = ½mv2

½u2 + gh = ½v2

u2 + 2gh = v2


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 = -10ms-1

Time up = time down or

Total time = 2 × time up


Six points

+Set positive

direction as up

v = 0

Use const accel equations with a = -10ms-1

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


Worked Examples


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 = 10ms-2

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.


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 = 10ms-2

v2 = u2 + 2ax

v2 = 2 × 10 × 8

v2 = 160

v = 12.6491

v 13 ms-1

+

Alternative – Energy Approach

Ug Ek

mgh = ½mv2

gh = ½v2

2gh = v2

2 × 10 × 8 = v2

160 = v2

12.6491 = vv = 13ms-1

• 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.


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 = 10ms-2

x = ut + ½ at2

x = ½ × 10 × 2.52

x = 31.25

x 31 s+


Example 3A stone is thrown down at 2.0ms-1 from a 3.0m tower and falls to the ground. What speed did the ground hit the ground?

v = ? u = 0x = 10m a = 10ms-2

v2 = u2 + 2ax

v2 = 22 + 2 × 10 × 3

v2 = 64

v = 8 ms-1

+ 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 = 8ms-1

2.0ms-1


Example 4A cannon ball is fired upwards from the ground at a speed of 30ms-1.(a) How high will the cannon ball reach?

x = ? u = 30ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax

0 = 302 + 2 × – 10 × x

0 = 900 + – 20 × x– 900 = – 20 × x

45 = x

x = 45m

+30ms-1

x


Example 4A cannon ball is fired upwards from the ground at a speed of 30ms-1.(a) How high will the cannon ball reach?

x = ? u = 30ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax

0 = 302 + 2 × – 10 × x

0 = 900 + – 20 × x– 900 = – 20 × x

45 = x

x = 45m

+30ms-1

x

Alternative – Energy Approach

EkUg ½mv2 = mgh

½v2 = ghv2 = 2gh

302 = 2 × 10 × h 900 = 20h

45 = hh = 45m


Example 4A cannon ball is fired upwards from the ground at a speed of 30ms-1.(b) What will be the flight time of the cannon ball?

Find time to top of the path

t = ? u = 30ms-1 v = 0 a = –10ms-2

v = u + at

0 = 30 + – 10 × t– 30 = – 10 × t

3 = t

Find total time

Total time = 2 × 3

= 6 seconds

+30ms-1

t t




t = ? u = 30ms-1 v = 0 a = –10ms-1

v = u + at

0 = 30 + – 10 × t– 30 = – 10 × t

3 = t

Find total time

Total time = 2 × 3

= 6 seconds

+30ms-1

t t

Alternative 1t = ? u = 30ms-1 x = 0 a = – 10ms-2

x = ut + ½at2

0 = 30t + ½ × – 10 × t2

0 = 30t + – 5 × t2

0 = 5t (6 – t)t = 0, 6So flight time = 6 seconds




t = ? u = 30ms-1 v = 0 a = –10ms-1

v = u + at

0 = 30 + – 10 × t– 30 = – 10 × t

3 = t

Find total time

Total time = 2 × 3

= 6 seconds

+30ms-1

t t

Alternative 2t = ? u = 30ms-1 v = – 30ms-1 a = – 10ms-2

v = u + at– 30 = 30 + – 10 × t– 60 = – 10 × t 6 = t

So flight time = 6 seconds


Example 4A cannon ball is fired upwards from the ground at a speed of 30ms-1.(c) What will be the speed at a height of 30m?

t = ? u = 30ms-1 x = 30m a = –

10ms-2

v2 = u2 + 2ax

v2 = 302 + 2 × – 10 × 30

v2 = 300

v = 17.321 ms-1

So the speed will be 17ms-1 at 30m on the

way up and 17ms-1 at 30m the way down

+30ms-1

30m

v

v


Example 4A cannon ball is fired upwards from the ground at a speed of 30ms-1.(c) What will be the speed at a height of 30m?

t = ? u = 30ms-1 x = 30m a = –

10ms-2

v2 = u2 + 2ax

v2 = 302 + 2 × – 10 × 30

v2 = 300

v = 17.321 ms-1

So the speed will be 17ms-1 at 30m on the

way up and 17ms-1 at 30m the way down

+30ms-1

30m

v

v

Alternative Using energy approach

v = ? u = 30ms-1 h = 30m g = 10ms-1

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 ms-1


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 = –10ms-2

x = ut + ½ at2

– 4 = u × 8 + ½ × – 10 × 82

– 4 = 8u – 320

316 = 8u

39.5 = u

u 40 ms-1

+

4.0m


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 = = = 34ms-2 v = u + atv = 0 + 34 × 1.5v = 51 ms-1

alternative using momentum/impulse approachv = ? t = 1.5s u = 0

Fnet = 17NFt = m(v – u)17 × 1.5 = 0.5v25.5 = 0.5v51 ms-1 = 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


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


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 = 10ms-1

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


u

Six points

y

x

Four landing formulae

On landing

thproj=

Rhproj = u

v =

= tan-1

vx

vyvx

vy

vx

vy

What do and u represent?

vx & vy

on landing

v

u


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


Example 1An aeroplane travelling at 360kmh-1 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 = 100ms-1 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 = 10ms-1

x = ut + ½ at2

300= ½ × 10 × t2

300 = 5 × t2

60 = t2

7.74597 = t

7.74597s

360kmh-1

= 100ms-1

y

x

d = ?


Example 1An aeroplane travelling at 360kmh-1 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 = 100ms-1 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 = 10ms-1

x = ut + ½ at2

300= ½ × 10 × t2

300 = 5 × t2

60 = t2

7.74597 = t

7.74597s

360kmh-1

= 100ms-1

y

x

Alternative using derived formulaeRhproj = ? u = 100ms-1 h = 300m g = 10ms-1

Rhproj = uRhproj = Rhproj = 774.597Rhproj 775m

d = ?



(b) What will be the velocity of the package when it hits the ground?

In y directionvy = ? uy = 0 x=300m a = 10ms-1

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 ms-1

360kmh-1

= 100ms-1

y

x

100ms-1

77.4597ms-1

tan = tan = = tan-1 = 37.761o

38o

v

38o

126 ms-1

v = ?




In y direction

vy = ? uy = 0 x=300m a = 10ms-1

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 ms-1

360kmh-1

= 100ms-1

y

x

100ms-1

77.4597ms-1

tan = tan = = tan-1 = 37.761o

38o

v38o

126 ms-1Alternative using derived formulaev = ? u = 100ms-1 h = 300m g = 10ms-1

v = v = v = 126.491v 126ms-1

= tan-1 = tan-1 = 37.761o 38o

38o

126 ms-1

v = ?




In y direction

vy = ? uy = 0 x=300m a = 10ms-1

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 ms-1

360kmh-1

= 100ms-1

y

x

100ms-1

tan = tan = = tan-1 = 37.761o

38o

v38o

126 ms-1

v = ?

Alternative if just asked for speed (energy approach)v = ? u = 100ms-1 h = 300m g = 10ms-1

Eki + Ug Ek

f

½mu2 + mgh = ½mv2

½u2 + gh = ½v2

u2 + 2gh = v2

1002 + 2×10×300 = v2

16000 = v2

126.491 = vv 126ms-1


Example 1(c) The aeroplane has to do another drop

but from a height 180m. What is the maximum speed (in kmh-1) 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 = 50ms-1

v = 180kmh-1

Find t from the y directiont = ? u = 0 x= 180m a = 10ms-1

x = ut + ½ at2

180= ½ × 10 × t2

180 = 5 × t2

36 = t2

6 = t

6s

v = kmh-1

y

x

d = 300m


Example 1(c) The aeroplane has to do another drop

but from a height 180m. What is the maximum speed (in kmh-1) 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 = 50ms-1

v = 180kmh-1


x = ut + ½ at2

180= ½ × 10 × t2

180 = 5 × t2

36 = t2

6 = t

6s

v = kmh-1

y

x

d = 300m

Alternative using derived formulaeu = ? Rhproj = 300 h = 180ms-1 g = 10ms-2

Rhproj = u300 = 300 = × 6 50 = uu = 180kmh-1


Example 2A car is driven off roof of a 15m building at 10ms-1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?

Find h2 from x direction

x =? u = 0 a = 10ms-2 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 = 10ms-1 d= 5.0m

t = t = t = 0.5 s

0.5s

y

x20ms-1

15mh1

h2

5.0m

Not drawn to scale


Example 2A car is driven off roof of a 15m building at 10ms-1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?

Find h2 from x direction

x =? u = 0 a = 10ms-2 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 = 10ms-1 d= 5.0m

t = t = t = 0.5 s

0.5s

y

x10ms-1

15mh1

h2

5.0m

Not drawn to scaleAlternative using derived formulaeh = h2 = ? Rhproj = 5.0 u = 10ms-1 g = 10ms-1

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 = 22.3607ms-1

v 22 ms-1


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 = 22.3607ms-1

v 22 ms-1


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 = 4.0ms-1 g = 10ms-2

Rhproj = u20 = 20 = × 0.89443

22.3607 = uu 22 ms-1

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 = 10ms-2

v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427

Find final velocity

h2 = a2 + b2 v2 = 252 + 8.9944272

v2 = 705v = 26.5518v 27 ms-1

25ms-1

8.94427ms-1

v

2002 Exam Q6




v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427

Find final velocity

h2 = a2 + b2 v2 = 252 + 8.9944272

v2 = 705v = 26.5518v 27 ms-1

25ms-1

8.94427ms-1

tan = tan = = tan-1 = 18.96173o

19o

v

19o

27 ms-1

tan = tan = = tan-1 = 37.761o

38o

v38o

126 ms-1Alternative using Derived Formulav = ? u = 25ms-1 h = 4.0m g = 10ms-1

v = v = v = 26.5518v 27ms-1

2002 Exam Q6




v2 = u2 + 2axv2 = 2 × 10 × 4v2 =80v = 8.94427

Find final velocity

h2 = a2 + b2 v2 = 252 + 8.9944272

v2 = 705v = 26.5518v 27 ms-1

25ms-1

8.94427ms-1

tan = tan = = tan-1 = 18.96173o

19o

v

19o

27 ms-1

tan = tan = = tan-1 = 37.761o

38o

v38o

126 ms-1Alternative using energy approachv = ? u = 25ms-1 h = 4.0m g = 10Nkg-1

Eki + Ug Ek

f

½mu2 + mgh = ½mv2

u2 + 2gh = v2

v =

v = v = 26.5518v 27ms-1

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 = = = 44ms-2 v = u + atv = 0 + 44 × 1.5v = 66 ms-1

In y directionv = ? t = 1.5s u = 0 a = 10ms-2 v = u + atv = 0 + 10 × 1.5v = 15 ms-1

alternative using momentum/impulse approachv = ? t = 1.5s u = 0

Fnet = 22NFt = m(v – u)22 × 1.5 = 0.5v33 = 0.5v66 ms-1 = 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 ms-1

66ms-1

15ms-1

tan = tan = = tan-1 = 12.8043o

13o

v

13o

68 ms-1

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 = –10ms-1

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


Worked Examples


Example 1A golf ball is hit from the ground at 31ms-1 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 = 28.096ms-1 t = ?

Find t to top of path from y dirnt = ? u = 13.101 v =0 a = –10ms-2

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

31ms-1

y

x

uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1

d

What are the three other ways of working out the flight time?1. t = ? u = 13.101 x =0 a = –10ms-2 (this will

involve a factorisation)2. t = ? u = 13.101 v = –13.101 a = –10ms-2

(this is pretty easy)3. Using flight time derived formula (just put the

numbers in and work out)




In x direction

d = ? v = 28.096ms-1 t = ?

Find t from y dirnt = ? u = 13.101 x =-0 a = –10ms-2

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 = 13.101ms-1

ux = 31cos25 = 28.096ms-1








In x direction

d = ? v = 28.096ms-1 t = ?

Find t from y dirnt = ? u = 13.101 v =-13.101 a = –10ms-2

v = u + at -13.101 = 13.101 + – 10 × t

– 26.202 = – 10 × t 2.6202 = t

y

x uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1








In x direction

d = ? v = 28.096ms-1 t = ?

d = v t

d = 28.096 × 2.6202

d = 73.617

d 74m

Find t from flight formulat = ? u = 31ms-1 = 25o g = –10ms-2

tflight =tflight =tflight = 2.6202

2.6202s

y

x uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1








In x direction

d = ? v = 28.096ms-1 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 = –10ms-2

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 = 31ms-1 = 25o g = 10ms-1

Robliq =Robliq =Robliq = 73.617Robliq 74m

25o

31ms-1

range



(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 28.096ms-1

25o

31ms-1

y

x

uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1



(c) How high will the golf ball rise?

In y directionx = ? u = 13.101ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax0 = 13.1012 + 2 × – 10 × x0 = 171.636 + – 20 × x

– 171.636 = – 20 × x 8.5818 = x

x = 8.6m

25o

31ms-1

y

x

uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1



(c) How high will the golf ball rise?

In y directionx = ? u = 13.101ms-1 v = 0 a = –10ms-2

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 = 31ms-1 = 25o g = 10ms-1

hmax= hmax=

hmax =8.5820hmax 8.6m

25o

31ms-1

y

x

uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1



(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

31ms-1

25o

31ms-1

y

x

uy = 31sin25 = 13.101ms-1

ux = 31cos25 = 28.096ms-1



(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 = 31ms-1 h = 20m g = 10ms-1

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 ms-1

25o

31ms-1

y

x


Example 2A golf ball is hit at 25ms-1 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 = 14.3390ms-1 t = 6.0s a = –10ms-2

x = ut + ½ at2

x = 14.3390 × 6 + ½ × – 10 × 62

x = – 93.966

so the cliff is 94m high

35o

25ms-1

y

x

uy = 25sin35 = 14.339ms-1

ux = 25cos35 = 20.4796ms-1

x


Example 2A golf ball is hit at 25ms-1 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 = 20.4796ms-1 t = 6

d = v t

d = 20.4796 × 6

d = 122.875

d 123m

35o

25ms-1

y

x

uy = 25sin35 = 14.339ms-1

ux = 25cos35 = 20.4796ms-1

d


Example 3A golf ball is hit at 28ms-1 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 = 24.249ms-1 a = –10ms-2 t = ?

x = ut + ½ at2

x = 24.249 × 4 + ½ × – 10 × 42

x = 16.996

so the ball hits the tree 17m up

60o

28ms-1

y

x

uy = 28sin60 = 24.249ms-1

ux = 28cos60 = 14ms-1

56m

h

Find t from the x directiont = ? v = ux = 14ms-1 d= 56m

t = t = t = 4 s

4s

28ms-1

60o


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 point-particle.) u = ? Robliq =10 = 20o g = 10ms-2

Robliq =

10 =

100 = u2sin40

155.572 = u2

12.4729 = u

u 12 ms-1

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 s-1.With what speed, relative to the deck, did the ball leave Fred's racket? Give your answer to three significant figures.

yx

8o

u

30ms-1

30ms-1

3.0m

8o

30ms-1

cos =

cos 8 =

u =

u = 30.29483ms-1

u 30.3ms-1

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 s-1.At its highest point, how far was the ball above the ground.

x = ? u = 30ms-1 v = 0 a = –10ms-2

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

30.29483ms-1

30ms-1

30ms-1

3.0m

8o

30ms-1

30.29483sin8= 4.216225ms-1

x

3.0m

2005 Exam Q12

Q3


x = ? u = 30ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax

0 = 4.2162252 + 2 × – 10 × x

0 = 17.77655 + – 20 × x– 17.77655 = – 20 × x

0.888828 = x


= 3.888828

3.89 m

yx

8o

30.29483ms-1

30ms-1

30ms-1

3.0m

8o

30ms-1

hmax

3.0m

Alternative using derived formulaehmax= ? u = 30.29483ms-1 = 8o g = 10ms-2

hmax= hmax=

hmax = 0.888828

2005 Exam Q12

Q3


x = ? u = 30ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax

0 = 4.2162252 + 2 × – 10 × x

0 = 17.77655 + – 20 × x– 17.77655 = – 20 × x

0.888828 = x


= 3.888828

3.89 m

yx

8o

30.29483ms-1

30ms-1

30ms-1

3.0m

8o

30ms-1

h2

Alternative using Energy Approachh2= ? u = 30.29483ms-1 h1 = 3m g = 10ms-2

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

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

Find t from the x directiont = ? v = ux = 28.096ms-1 d= 127m

t = t = t = 3.5032 st 3.5 s

2007 Exam Q14


25o

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

AlternativeFind t from the y directiont = ? u = 16.9047ms-1 v = –16.9047ms-1 a = –10ms-2

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


25o

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

Find t from the y directiont = ? u = 16.9047ms-1 v = –16.9047ms-1 a = –10ms-2

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


25o

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

Alternative with formulaFind t from flight formula t = ? u = 40ms-1 = 25o g = –10ms-2

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

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

Find hmax from y direction x = ? u = 16.9047ms-1 v = 0 a = –

10ms-2 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

40ms-1y

x uy = 40sin25 = 16.9047ms-1

ux = 40cos25 = 36.2523ms-1

Alternative with formulaFind t from flight formula

hmax= ? u = 40ms-1 = 25o g = 10ms-2

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)

Projectile Motion Notes. Vertical Projectile Motion.

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