Projectile Motion Notes. Vertical Projectile Motion.

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Projectile Motion NotesVertical Projectile MotionVertical Projectile MotionCase 1 Case 2 Case 3 Case 4 uuuObject dropped from restObject projected downwardsObject launched vertically and returns to launch heightObject launched vertically and returns to a different height3Case 1: Vertical Projectile Motion+Three pointsSet positive direction as downIf time not involved can use energy approachUg Ek mgh = mv2gh = v22gh = v2Use constant acceleration equations with u = 0 a = 10ms-1Case 2: Vertical Projectile Motion+Three pointsSet positive direction as downUse constant acceleration equations with a = 10ms-1uIf time not involved can use energy approachEki + Ug Ekf mu2 + mgh = mv2 u2 + gh = v2 u2 + 2gh = v2Case 3: Vertical Projectile Motion Six points+Set positive direction as upuSpeeds up are the same as down at each heightvvv = 0If time not involved can use energy approachEki Ekf + Ug mu2 = mv2 + mghu2 = v2 + ghu2 = v2 + 2ghUse const accel equations with a = -10ms-1Time up = time down or Total time = 2 time upCase 4: Vertical Projectile MotionSix points+Set positive direction as upv = 0Use const accel equations with a = -10ms-1Speeds up are the same as down at each heightIf time not involved can use energy approachEki Ekf + Ug mu2 = mv2 + mghDisplacements below launch height will be negativeuCase 4: Vertical Projectile MotionSix points+Set positive direction as upv = 0Use const accel equations with a = -10ms-1Speeds up are the same as down at each heightIf time not involved can use energy approachEki Ekf + Ug mu2 = mv2 + mghDisplacements below launch height will be negativeuVertical Projectile MotionWorked ExamplesVertical Projectile MotionExample 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 = 8.0ma = 10ms-2 x = ut + at28 = 10 t28 = 5 t21.6 = t21.26491 = tt 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 answersIn 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 MotionExample 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 = 8.0ma = 10ms-2 v2 = u2 + 2axv2 = 2 10 8v2 = 160 v = 12.6491 v 13 ms-1

+Alternative Energy ApproachUg Ek mgh = mv2gh = v22gh = v22 10 8 = v2160 = v212.6491 = vv = 13ms-1In 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 MotionExample 2A stone is dropped down a well and takes 2.5 seconds to hit the water. How deep is the well?

x = ? u = 0t = 2.5sa = 10ms-2 x = ut + at2x = 10 2.52x = 31.25x 31 s

+Vertical Projectile MotionExample 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 = 10ma = 10ms-2 v2 = u2 + 2axv2 = 22 + 2 10 3v2 = 64 v = 8 ms-1

+Alternative Energy ApproachEki + Ug Ekf mu2 + mgh = mv2 u2 + gh = v2 u2 + 2gh = v222 + 2 10 3 = v264 = v28 = vv = 8ms-12.0ms-1Vertical Projectile MotionExample 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 = 0a = 10ms-2

v2 = u2 + 2ax0 = 302 + 2 10 x0 = 900 + 20 x 900 = 20 x 45 = xx = 45m

+30ms-1xVertical Projectile MotionExample 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 = 0a = 10ms-2

v2 = u2 + 2ax0 = 302 + 2 10 x0 = 900 + 20 x 900 = 20 x 45 = xx = 45m

+30ms-1xAlternative Energy ApproachEkUg mv2 = mghv2 = ghv2 = 2gh302 = 2 10 h 900 = 20h45 = hh = 45mVertical Projectile MotionExample 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 = 0a = 10ms-2 v = u + at0 = 30 + 10 t 30 = 10 t 3 = tFind total time Total time = 2 3 = 6 seconds

+30ms-1ttVertical Projectile MotionExample 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 = 0a = 10ms-1 v = u + at0 = 30 + 10 t 30 = 10 t 3 = tFind total time Total time = 2 3 = 6 seconds

+30ms-1ttAlternative 1t = ? u = 30ms-1 x = 0 a = 10ms-2x = ut + at20 = 30t + 10 t20 = 30t + 5 t20 = 5t (6 t)t = 0, 6So flight time = 6 seconds

Vertical Projectile MotionExample 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 = 0a = 10ms-1 v = u + at0 = 30 + 10 t 30 = 10 t 3 = tFind total time Total time = 2 3 = 6 seconds

+30ms-1ttAlternative 2t = ? u = 30ms-1 v = 30ms-1 a = 10ms-2 v = u + at 30 = 30 + 10 t 60 = 10 t 6 = tSo flight time = 6 seconds

Vertical Projectile MotionExample 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 = 30ma = 10ms-2 v2 = u2 + 2axv2 = 302 + 2 10 30v2 = 300v = 17.321 ms-1So the speed will be 17ms-1 at 30m on the way up and 17ms-1 at 30m the way down

+30ms-130mvvVertical Projectile MotionExample 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 = 30ma = 10ms-2 v2 = u2 + 2axv2 = 302 + 2 10 30v2 = 300v = 17.321 ms-1So the speed will be 17ms-1 at 30m on the way up and 17ms-1 at 30m the way down

+30ms-130mvvAlternative Using energy approach v = ? u = 30ms-1 h = 30m g = 10ms-1Eki Ekf + Ug mu2 = mv2 + mgh u2 = v2+ gh u2 = v2+ 2gh 302 = v2+ 21030 900 = v2 + 600300 = v2 17.3205 = vv 17 ms-1Vertical Projectile MotionExample 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 + at2 4 = u 8 + 10 82 4 = 8u 320 316 = 8u39.5 = uu 40 ms-1

+4.0mVertical Projectile MotionExam Questions2006 ExamQ6 & 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?

+22NW = mg = 0.5 10 = 5NWhen 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 problemHorizontally Launched Projectile Motion

Horizontally Launched Projectile MotionCase 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 aircraftu

u

uHorizontally Launched Projectile MotionCase 1 Case 2 u

uHorizontally launched object drops to another heightEx object launched of building to the top of another building

Case 2: Horizontally Launched Projectile Motion uSix pointsReference axesyxIn y direction uy = 0 a = 10ms-1and use const accel equnsIf time not involvedEki + Ug Ekf mu2 + mgh = mv2u2 + 2gh = v2In x dirn vx = u and usedvtIf not enough information in one direction get time from the other directionvxvyvxvyvxvyThe motion in the x & y directions are independentvy is accelerated by gvx is constant = u

Case 2: Horizontally Launched Projectile Motion uSix pointsyxFour landing formulaeOn landingvxvyvxvyvxvyvu

Case 2: Horizontally Launched Projectile Motion Six pointsIf you treat the top of the second building as a virtual ground level all the previous five points including formulae hold for this situationh1h2h = h1 h2Virtual ground levelFour landing formulaeHorizontal Projectile MotionWorked ExamplesHorizontally Launched Projectile MotionExample 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 directiond = ? v = ux = 100ms-1 t = ?d = v td = 100 7.74597d = 774.597d 775m

Find t from the y directiont = ? u = 0 x=8.0m a = 10ms-1 x = ut + at2300= 10 t2300 = 5 t260 = t27.74597 = t7.74597s360kmh-1= 100ms-1

yxd = ?Horizontally Launched Projectile MotionExample 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 directiond = ? v = ux = 100ms-1 t = ?d = v td = 100 7.74597d = 774.597d 775m

Find t from the y directiont = ? u = 0 x=8.0m a = 10ms-1 x = ut + at2300= 10 t2300 = 5 t260 = t27.74597 = t7.74597s360kmh-1= 100ms-1

yxd = ?Horizontally Launched Projectile MotionExample 1An aeroplane travelling at 360kmh-1 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 = 10ms-1 v2 = u2 + 2axv2 = 2 10 300v2 =6000 v = 77.4597

Find final velocity

h2 = a2 + b2 v2 = 1002 + 77.45072v2 = 16000v = 126.49v 126 ms-1360kmh-1= 100ms-1

yx100ms-177.4597ms-1v38o126 ms-1v = ?Horizontally Launched Projectile MotionExample 1An aeroplane travelling at 360kmh-1 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 = 10ms-1 v2 = u2 + 2axv2 = 2 10 300v2 =6000 v = 77.4597

Find final velocity

h2 = a2 + b2 v2 = 1002 + 77.45072v2 = 16000v = 126.49v 126 ms-1360kmh-1= 100ms-1

yx100ms-177.4597ms-1v38o126 ms-138o126 ms-1v = ?Horizontally Launched Projectile MotionExample 1An aeroplane travelling at 360kmh-1 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 = 10ms-1 v2 = u2 + 2axv2 = 2 10 300v2 =6000 v = 77.4597

Find final velocity

h2 = a2 + b2 v2 = 1002 + 77.45072v2 = 16000v = 126.49v 126 ms-1360kmh-1= 100ms-1

yx100ms-1v38o126 ms-1v = ?Alternative if just asked for speed (energy approach)v = ? u = 100ms-1 h = 300m g = 10ms-1Eki + Ug Ekf mu2 + mgh = mv2u2 + gh = v2u2 + 2gh = v21002 + 210300 = v216000 = v2126.491 = vv 126ms-1

Horizontally Launched Projectile MotionExample 1Find t from the y directiont = ? u = 0 x= 180m a = 10ms-1 x = ut + at2180= 10 t2180 = 5 t236 = t26 = t6sv = kmh-1

yxd = 300mHorizontally Launched Projectile MotionExample 1Find t from the y directiont = ? u = 0 x= 180m a = 10ms-1 x = ut + at2180= 10 t2180 = 5 t236 = t26 = t6sv = kmh-1

yxd = 300mHorizontally Launched Projectile MotionExample 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 directionx =? u = 0 a = 10ms-2 t = ?x = ut + at2x = 10 0.52x = 1.25mFind height of second buildingHeight = 15 1.25 = 13.75 14m0.5syx20ms-1

15mh1h25.0mNot drawn to scaleHorizontally Launched Projectile MotionExample 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 directionx =? u = 0 a = 10ms-2 t = ?x = ut + at2x = 10 0.52x = 1.25mFind height of second buildingHeight = 15 1.25 = 13.75 14m0.5syx10ms-1

15mh1h25.0mNot drawn to scaleHorizontal Projectile MotionExam Questions2002 Exam Q5Q1A 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 .

Find t from the y directiont = ? u = 0 x= 4m a = 10ms-2 x = ut + at24= 10 t24 = 5 t20.8 = t20.89443 = t0.89443s2002 Exam Q5Q1A 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 .

Find t from the y directiont = ? u = 0 x= 4m a = 10ms-2 x = ut + at24= 10 t24 = 5 t20.8 = t20.89443 = t0.89443s2002 Exam Q6Q2In 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 ms1Calculate 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 + 8.9944272v2 = 705v = 26.5518v 27 ms-1 25ms-18.94427ms-1v2002 Exam Q6Q2In 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 ms1Calculate 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 + 8.9944272v2 = 705v = 26.5518v 27 ms-1 25ms-18.94427ms-1v19o 27 ms-1v38o126 ms-12002 Exam Q6Q2In 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 ms1Calculate 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 + 8.9944272v2 = 705v = 26.5518v 27 ms-1 25ms-18.94427ms-1v19o 27 ms-1v38o126 ms-1Same as derived formula2006 Exam Q8Q3A 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?

22Nyx

alternative using momentum/impulse approachv = ? t = 1.5s u = 0 Fnet = 22NFt = m(v u)22 1.5 = 0.5v33 = 0.5v66 ms-1 = v2006 Exam Q8Q3After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?

22Nyx

Find final velocity

h2 = a2 + b2 v2 = 662 + 152v2 = 4581v = 67.6831v 68 ms-1 66ms-115ms-1v13o 68 ms-1Obliquely Launched Projectile MotionObliquely Launched Projectile MotionCase 1: landing height = launch heightCase 2: landing height different to launch height

Case 1: Obliquely Launched Projectiles Landing At The Launching Heightyxuuy = u sin ux = u cos Reference axesIn y direction uy = u sin a = 10ms-1 use const accel equnsIn x dirn vx = u cos in If not enough info in one dirn get time from the other dirnThe motion in the x & y directions are independentuy = u sin vx is constant = u cos

dvtvxvyvxvyvxEight pointsIf time not involvedEki Ekf + Ug mu2 = mv2+ mgh u2 = v2+ 2gh vyv = u=vxSpeeds before hmax are the same as those after at each heightEkmin = mvx2

Case 1: Obliquely Launched Projectiles Landing At The Launching Heightyxuuy = u sin ux = u cos vxvyvxvxvyvxvyv = u=Three formulae

Case 2: Obliquely Launched Projectiles Landing At A Different Height To The Launching Heightu

x

Obliquely Launched Projectile MotionWorked ExamplesObliquely Launched Projectile MotionExample 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 directiond = ? 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 = tFind total time Total time = 2 1.3101 = 2.6202s25o31ms-1yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1dWhat are the three other ways of working out the flight time?t = ? u = 13.101 x =0 a = 10ms-2 (this will involve a factorisation)t = ? u = 13.101 v = 13.101 a = 10ms-2 (this is pretty easy)Using flight time derived formula (just put the numbers in and work out) Obliquely Launched Projectile MotionExample 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 directiond = ? v = 28.096ms-1 t = ?

yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1What are the three other ways of working out the flight time?t = ? u = 13.101 x =0 a = 10ms-2 (this will involve a factorisation)t = ? u = 13.101 v = 13.101 a = 10ms-2 (this is pretty easy)Using flight time derived formula (just put the numbers in and work out) Obliquely Launched Projectile MotionExample 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 directiond = ? 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 = tyxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1What are the three other ways of working out the flight time?t = ? u = 13.101 x =0 a = 10ms-2 (this will involve a factorisation)t = ? u = 13.101 v = 13.101 a = 10ms-2 (this is pretty easy)Using flight time derived formula (just put the numbers in and work out) Obliquely Launched Projectile MotionExample 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 directiond = ? v = 28.096ms-1 t = ?d = v td = 28.096 2.6202d = 73.617d 74m

2.6202syxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1What are the three other ways of working out the flight time?t = ? u = 13.101 x =0 a = 10ms-2 (this will involve a factorisation)t = ? u = 13.101 v = 13.101 a = 10ms-2 (this is pretty easy)Using flight time derived formula (just put the numbers in and work out) Obliquely Launched Projectile MotionExample 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 directiond = ? v = 28.096ms-1 t = ?d = v td = 28.096 2.6202d = 73.617d 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 = tFind total time Total time = 2 1.3101 = 2.6202s2.6202s25o31ms-1rangeObliquely Launched Projectile MotionExample 1A golf ball is hit from the ground at 31ms-1 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 28.096ms-1

25o31ms-1yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1Obliquely Launched Projectile MotionExample 1A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.(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 = xx = 8.6m 25o31ms-1yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1Obliquely Launched Projectile MotionExample 1A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.(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 = xx = 8.6m 25o31ms-1yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1Obliquely Launched Projectile MotionExample 1A golf ball is hit from the ground at 31ms-1 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:

25o31ms-125o31ms-1yxuy = 31sin25 = 13.101ms-1ux = 31cos25 = 28.096ms-1Obliquely Launched Projectile MotionExample 1A golf ball is hit from the ground at 31ms-1 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 = 31ms-1 h = 20m g = 10ms-1Eki Ekf + Ug mu2 = mv2 + mgh u2 = v2+ gh u2 = v2+ 2gh 312 = v2+ 21020 961 = v2 + 400561 = v2 23.685 = vv 24 ms-125o31ms-1yxObliquely Launched Projectile MotionExample 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 + at2x = 14.3390 6 + 10 62x = 93.966so the cliff is 94m high35o25ms-1yxuy = 25sin35 = 14.339ms-1ux = 25cos35 = 20.4796ms-1xObliquely Launched Projectile MotionExample 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 = 6d = v td = 20.4796 6d = 122.875d 123m35o25ms-1yxuy = 25sin35 = 14.339ms-1ux = 25cos35 = 20.4796ms-1dObliquely Launched Projectile MotionExample 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 + at2x = 24.249 4 + 10 42x = 16.996so the ball hits the tree 17m up60o28ms-1yxuy = 28sin60 = 24.249ms-1ux = 28cos60 = 14ms-156mh4s28ms-160oObliquely Launched Projectile MotionExam Questions2004 Exam Q7Q1

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 Q11Q2 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.

yx8ou30ms-130ms-13.0m8o30ms-12005 Exam Q12Q3 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 + 2ax0 = 4.2162252 + 2 10 x0 = 17.77655 + 20 x 17.77655 = 20 x 0.888828 = xOverall height = 0.888828 + 3 = 3.888828 3.89 m

yx8o30.29483ms-130ms-130ms-13.0m8o30ms-130.29483sin8= 4.216225ms-1x3.0m2005 Exam Q12Q3 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 + 2ax0 = 4.2162252 + 2 10 x0 = 17.77655 + 20 x 17.77655 = 20 x 0.888828 = xOverall height = 0.888828 + 3 = 3.888828 3.89 m

yx8o30.29483ms-130ms-130ms-13.0m8o30ms-1hmax3.0m2005 Exam Q12Q3 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 + 2ax0 = 4.2162252 + 2 10 x0 = 17.77655 + 20 x 17.77655 = 20 x 0.888828 = xOverall height = 0.888828 + 3 = 3.888828 3.89 m

yx8o30.29483ms-130ms-130ms-13.0m8o30ms-1h2Alternative using Energy Approachh2= ? u = 30.29483ms-1 h1 = 3m g = 10ms-2 Eki + Ugi Ekf + Ugf mu2 + mgh1 = mv2 + mgh2 u2 + gh1 = v2 + gh2 30.294832 + 10 3 = 302 + 10 h2 488.888 = 450 + 10 h2 3.8888 = h2 h2 3.89m

h1 =2007 Exam Q14Q4Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the time of ight of the paintball?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-12007 Exam Q14Q4Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the time of ight of the paintball?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-1AlternativeFind 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 sThere is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction2007 Exam Q14Q4Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the time of ight of the paintball?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-1Find 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 sThere is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction2007 Exam Q14Q4Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the time of ight of the paintball?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-1There is a discrepancy in the data for this problem hence the slight difference in answers from the previous methods2007 Exam Q15Q5Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the value of h, the maximum height above the ring level?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-1Find 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 = xx 14m

2007 Exam Q15Q5Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.What is the value of h, the maximum height above the ring level?

25o40ms-1yxuy = 40sin25 = 16.9047ms-1ux = 40cos25 = 36.2523ms-1There is a discrepancy in the data for this problem hence the slight difference in answers from the previous method2007 Exam Q16Q6Daniel res a paintball at an angle of 25 to the horizontal and a speed of 40.0 ms1. 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 red are the same.Which of the following diagrams (AD) 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 Q17Q7Later in the game, Daniel is twice as far away from John (254 m). John res 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.ight time B.initial speed C.accelerationD.angle of ring 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 ProblemSome Loose Rules For Deciding On The Best Method To Solve A ProblemAn energy approach is very useful when asked for the magnitude of a velocity that is not at launching height or maximum height.Remember there are three approaches that can be used for solving Constant Accel EquationsImpulse/ Momentum Formulae(no dist)Energy Formulae(no time)