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Projectile Motion Notes

Projectile Motion Notes. Vertical Projectile Motion. Vertical Projectile Motion. Case 1 Case 2 Case 3 Case 4 . Object launched vertically and returns to launch height. Object launched vertically and returns to a different height. u. u. Object projected downwards.

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Projectile Motion Notes

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  1. Projectile Motion Notes

  2. Vertical Projectile Motion

  3. Vertical Projectile Motion Case 1 Case 2 Case 3 Case 4 Object launched vertically and returns to launch height Object launched vertically and returns to a different height u u Object projected downwards Object dropped from rest u

  4. Case 1: Vertical Projectile Motion Set positive direction as down + Use constant acceleration equations with u = 0 a = 10ms-1 If time not involved can use energy approach Ug Ek mgh = ½mv2 gh= ½v2 2gh= v2 Three points

  5. Case 2: Vertical Projectile Motion Set positive direction as down + Use constant acceleration equations with a = 10ms-1 u If time not involved can use energy approach Eki+ Ug Ekf ½mu2 + mgh = ½mv2 ½u2+ gh= ½v2 u2+ 2gh= v2 Three points

  6. Case 3: Vertical Projectile Motion If time not involved can use energy approach Eki Ekf + Ug ½mu2 = ½mv2 + mgh ½u2 = ½v2 + gh u2 = v2 + 2gh v = 0 + Speeds up are the same as down at each height Use constaccel equations with a = -10ms-1 u v v Time up = time down or Total time = 2 × time up Set positive direction as up Six points

  7. Case 4: Vertical Projectile Motion If time not involved can use energy approach Eki Ekf + Ug ½mu2 = ½mv2 + mgh v = 0 + Speeds up are the same as down at each height Use constaccel equations with a = -10ms-1 u Displacements below launch height will be negative Set positive direction as up Six points

  8. Case 4: Vertical Projectile Motion If time not involved can use energy approach Eki Ekf + Ug ½mu2 = ½mv2 + mgh v = 0 + Speeds up are the same as down at each height Use constaccel equations with a = -10ms-1 u Displacements below launch height will be negative Set positive direction as up Six points

  9. Vertical Projectile Motion Worked Examples

  10. Vertical Projectile MotionExample 1 A 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 = 0 x = 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 eithernot relevant orit represents a direction which is already obvious in the problem. In this case a negative time is not plausible.

  11. Vertical Projectile MotionExample 1 A stone is dropped from an 8.0m tower and falls to the ground. (b)What speed will the stone hit the ground? v = ? u = 0 x = 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 = v v = 13ms-1 • In In Physics we normally just write out the magnitude of the square root because the negative value is eithernot relevant orit 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.

  12. Vertical Projectile MotionExample 2 A 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 +

  13. Vertical Projectile MotionExample 3 A 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 = 0 x = 10m a = 10ms-2 v2 = u2 + 2ax v2= 22+ 2 × 10 × 3 v2 = 64 v = 8 ms-1 + Alternative – Energy Approach Eki+ Ug Ekf ½mu2 + mgh = ½mv2 ½u2 + gh = ½v2 u2 + 2gh = v2 22+ 2 × 10 × 3 = v2 64 = v2 8 = v v = 8ms-1 2.0ms-1

  14. Vertical Projectile MotionExample 4 A 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-1v= 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

  15. Vertical Projectile MotionExample 4 A 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-1v= 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 Alternative – Energy Approach EkUg ½mv2 = mgh ½v2 = gh v2 = 2gh 302 = 2 × 10 × h 900 = 20h 45 = h h = 45m x

  16. Vertical Projectile MotionExample 4 A 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-1v= 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

  17. Vertical Projectile MotionExample 4 A 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-1v= 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 Alternative 1 t = ? u = 30ms-1x = 0 a = – 10ms-2 x = ut + ½at2 0= 30t+ ½ × – 10 × t2 0= 30t + – 5 × t2 0= 5t (6 – t) t= 0, 6 So flight time = 6 seconds + 30ms-1 t t

  18. Vertical Projectile MotionExample 4 A 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-1v= 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 Alternative 2 t = ? u = 30ms-1v = – 30ms-1 a = – 10ms-2 v= u + at – 30= 30 + – 10 × t – 60 = – 10 × t 6 = t So flight time = 6 seconds + 30ms-1 t t

  19. Vertical Projectile MotionExample 4 A 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-1x= 30ma = –10ms-2 v2 = u2 + 2ax v2= 302 + 2 × – 10 × 30 v2 = 300 v=  17.321 ms-1 So the speed will be 17ms-1at 30m on the way up and 17ms-1at 30m the way down + 30ms-1 v v 30m

  20. Vertical Projectile MotionExample 4 A 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-1x= 30ma = –10ms-2 v2 = u2 + 2ax v2= 302 + 2 × – 10 × 30 v2 = 300 v=  17.321 ms-1 So the speed will be 17ms-1at 30m on the way up and 17ms-1at 30m the way down Alternative Using energy approach v = ? u = 30ms-1h = 30m g = 10ms-1 Eki Ekf+ Ug ½mu2= ½mv2 + mgh ½u2= ½v2+ gh u2= v2+ 2gh 302= v2+ 2×10×30 900 = v2 + 600 300 = v2 17.3205 = v v  17ms-1 + 30ms-1 v v 30m

  21. Vertical Projectile MotionExample 5 If 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

  22. Vertical Projectile Motion Exam Questions

  23. 2006 ExamQ6 & 7 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 A 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 v= ? t= 1.5su = 0 a= = = 34ms-2 v = u + at v= 0 + 34 × 1.5 v = 51 ms-1 alternative using momentum/impulse approach v = ? t = 1.5su = 0 Fnet = 17N Ft= m(v – u) 17 × 1.5 = 0.5v 25.5 = 0.5v 51 ms-1 = v W = mg = 0.5× 10 = 5N

  24. Horizontally Launched Projectile Motion

  25. Horizontally Launched Projectile Motion Case 1 Case 2 Horizontally launched object drops to ground level Ex1 – object launched of building/cliff Ex2 – object falling off a moving vehicle Ex3 – object dropped from an aircraft u u u

  26. Horizontally Launched Projectile Motion Case 1 Case 2 Horizontally launched object drops to another height Ex – object launched of building to the top of another building u u

  27. Case 2: Horizontally Launched Projectile Motion Reference axes The motion in the x & y directions are independent vyis accelerated by gvxis constant = u In y direction uy = 0 a = 10ms-1 and use constaccelequns x In x dirnvx= u and use u vx vx vx If not enough information in one direction get time from the other direction d vy vy t v y vy If time not involved Eki+ Ug Ekf ½mu2 + mgh = ½mv2 u2+ 2gh= v2 Six points

  28. Case 2: Horizontally Launched Projectile Motion On landing thproj= x Rhproj = u u vx vx vx v = What do and u represent? vx & vy on landing vy vy y vy •  = tan-1  Four landing formulae Six points v

  29. Case 2: Horizontally Launched Projectile Motion 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 h2 h = h1 – h2 h1 u Virtual ground level Four landing formulae Six points

  30. Horizontal Projectile Motion Worked Examples

  31. Horizontally Launched Projectile MotionExample 1 360kmh-1 = 100ms-1 An 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-1t = ? d = vt d = 100 × 7.74597 d = 774.597 d  775m x d = ? Find t from the y direction t = ? 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 y

  32. Horizontally Launched Projectile MotionExample 1 360kmh-1 = 100ms-1 An 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-1t = ? d = vt d = 100 × 7.74597 d = 774.597 d  775m x d = ? Alternative using derived formulae Rhproj = ? u = 100ms-1h = 300m g = 10ms-1 Rhproj = u Rhproj= Rhproj= 774.597 Rhproj 775m Find t from the y direction t = ? 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 y

  33. Horizontally Launched Projectile MotionExample 1 360kmh-1 = 100ms-1 An 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 direction vy = ? uy= 0 x=300m a = 10ms-1 v2 = u2 + 2ax v2 = 2 × 10 × 300 v2=6000 v = 77.4597 x Find final velocity h2 = a2 + b2 v2= 1002+ 77.45072 v2= 16000 v= 126.49 v 126 ms-1 v = ? 100ms-1  77.4597ms-1 v tan  = tan  =  = tan-1  = 37.761o   38o y 38o 126 ms-1

  34. Horizontally Launched Projectile MotionExample 1 360kmh-1 = 100ms-1 An 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 direction vy = ? uy= 0 x=300m a = 10ms-1 v2 = u2 + 2ax v2 = 2 × 10 × 300 v2=6000 v = 77.4597 x Find final velocity h2 = a2 + b2 v2= 1002+ 77.45072 v2= 16000 v= 126.49 v 126 ms-1 v = ? 100ms-1  38o Alternative using derived formulae v = ? u = 100ms-1h = 300m g = 10ms-1 v = v = v = 126.491 v  126ms-1 77.4597ms-1 38o v 126 ms-1 126 ms-1 y •  = tan-1 •  = tan-1 •  = 37.761o •  38o tan  = tan  =  = tan-1  = 37.761o   38o

  35. Horizontally Launched Projectile MotionExample 1 360kmh-1 = 100ms-1 An 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 direction vy = ? uy= 0 x=300m a = 10ms-1 v2 = u2 + 2ax v2 = 2 × 10 × 300 v2=6000 v = 77.4597 x v = ? Find final velocity h2 = a2 + b2 v2= 1002+ 77.45072 v2= 16000 v= 126.49 v 126 ms-1 Alternative if just asked for speed (energy approach) v = ? u = 100ms-1h = 300m g = 10ms-1 Eki+ Ug Ekf ½mu2 + mgh = ½mv2 ½u2+ gh= ½v2 u2 + 2gh = v2 1002+ 2×10×300 = v2 16000 = v2 126.491 = v v  126ms-1 100ms-1  38o v 126 ms-1 y tan  = tan  =  = tan-1  = 37.761o   38o

  36. Horizontally Launched Projectile MotionExample 1 v = kmh-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= 300mt = ? v= v= v = 50ms-1 v = 180kmh-1 x d = 300m Find t from the y direction t = ? u= 0 x= 180m a= 10ms-1 x= ut + ½ at2 180= ½ × 10 × t2 180 = 5 × t2 36 = t2 6 = t y 6s

  37. Horizontally Launched Projectile MotionExample 1 v = kmh-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= 300mt = ? v= v= v = 50ms-1 v = 180kmh-1 x d = 300m Alternative using derived formulae u = ? Rhproj = 300 h = 180ms-1g = 10ms-2 Rhproj = u 300= 300= × 6 50 = u u = 180kmh-1 Find t from the y direction t = ? u= 0 x= 180m a= 10ms-1 x= ut + ½ at2 180= ½ × 10 × t2 180 = 5 × t2 36 = t2 6 = t y 6s

  38. Horizontally Launched Projectile MotionExample 2 A 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-2t = ? x= ut + ½ at2 x = ½ × 10 × 0.52 x= 1.25m Find height of second building Height = 15 – 1.25 = 13.75  14m x h1 h2 Not drawn to scale 15m 5.0m 20ms-1 0.5s Find t from the xdirection t = ? v= ux= 10ms-1d= 5.0m t= t= t = 0.5s y

  39. Horizontally Launched Projectile MotionExample 2 A 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-2t = ? x= ut + ½ at2 x = ½ × 10 × 0.52 x= 1.25m Find height of second building Height = 15 – 1.25 = 13.75  14m x h1 h2 Not drawn to scale 15m 5.0m Alternative using derived formulae h = h2 = ? Rhproj = 5.0 u = 10ms-1g = 10ms-1 Rhproj = u 5 = 0.5= 10ms-1 0.5s Find t from the xdirection t = ? v= ux= 10ms-1d= 5.0m t= t= t = 0.5s y 0.25 = 1.25 = h Find height of second building Height = 15 – 1.25 = 13.75  14m

  40. Horizontal Projectile Motion Exam Questions

  41. 2002 Exam Q5Q1 A 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 direction v = ux= ? d= 20mt= ? v= v= v = 22.3607ms-1 v  22 ms-1 0.89443s Find t from the y direction t = ? u= 0 x= 4m a= 10ms-2 x= ut + ½ at2 4= ½ × 10 × t2 4= 5 × t2 0.8 = t2 0.89443 = t

  42. 2002 Exam Q5Q1 A 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 . Alternative using derived formulae u = ? Rhproj = 20.0 h = 4.0ms-1g = 10ms-2 Rhproj = u 20= 20= × 0.89443 22.3607 = u u 22 ms-1 In x direction v = ux= ? d= 20mt= ? v= v= v = 22.3607ms-1 v  22 ms-1 0.89443s Find t from the y direction t = ? u= 0 x= 4m a= 10ms-2 x= ut + ½ at2 4= ½ × 10 × t2 4= 5 × t2 0.8 = t2 0.89443 = t

  43. 2002 Exam Q6Q2 In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1 Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2. Find final velocity h2 = a2 + b2 v2= 252+ 8.9944272 v2= 705 v= 26.5518 v 27 ms-1 25ms-1 8.94427ms-1 v In y direction vy = ? uy= 0 x=4.0m a = 10ms-2 v2 = u2 + 2ax v2 = 2 × 10 × 4 v2=80 v = 8.94427

  44. 2002 Exam Q6Q2 In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1 Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2. Find final velocity h2 = a2 + b2 v2= 252+ 8.9944272 v2= 705 v= 26.5518 v 27 ms-1 25ms-1  8.94427ms-1  38o Alternative using Derived Formula v = ? u = 25ms-1h = 4.0m g = 10ms-1 v = v = v = 26.5518 v  27ms-1 v tan  = tan  =  = tan-1  = 18.96173o   19o v 126 ms-1 In y direction vy = ? uy= 0 x=4.0m a = 10ms-2 v2 = u2 + 2ax v2 = 2 × 10 × 4 v2=80 v = 8.94427 tan  = tan  =  = tan-1  = 37.761o   38o 19o 27 ms-1

  45. 2002 Exam Q6Q2 In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1 Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2. Find final velocity h2 = a2 + b2 v2= 252+ 8.9944272 v2= 705 v= 26.5518 v 27 ms-1 25ms-1  8.94427ms-1  38o Alternative using energy approach v = ? u = 25ms-1h = 4.0m g = 10Nkg-1 Eki+ Ug Ekf ½mu2 + mgh = ½mv2 u2+ 2gh = v2 v = v tan  = tan  =  = tan-1  = 18.96173o   19o v 126 ms-1 In y direction vy = ? uy= 0 x=4.0m a = 10ms-2 v2 = u2 + 2ax v2 = 2 × 10 × 4 v2=80 v = 8.94427 v = v = 26.5518 v  27ms-1 tan  = tan  =  = tan-1  = 37.761o   38o 19o Same as derived formula 27 ms-1

  46. 2006 Exam Q8Q3 A 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? x In x direction v= ? t= 1.5su = 0 a= = = 44ms-2 v = u + at v= 0 + 44 × 1.5 v = 66 ms-1 In y direction v = ? t = 1.5su = 0 a = 10ms-2 v = u + at v = 0 + 10 × 1.5 v = 15 ms-1 22N y alternative using momentum/impulse approach v = ? t = 1.5su = 0 Fnet = 22N Ft= m(v – u) 22 × 1.5 = 0.5v 33 = 0.5v 66 ms-1 = v

  47. 2006 Exam Q8Q3 After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground? x Find final velocity h2 = a2 + b2 v2= 662+ 152 v2= 4581 v= 67.6831 v 68 ms-1 66ms-1  15ms-1 22N v tan  = tan  =  = tan-1  = 12.8043o   13o y 13o 68 ms-1

  48. Obliquely Launched Projectile Motion

  49. Obliquely Launched Projectile Motion Case 1: landing height = launch height Case 2: landing height different to launch height

  50. Case 1: Obliquely Launched Projectiles Landing At The Launching Height In x dirn vx= u cosin In y direction uy = u sina = –10ms-1use constaccelequns The motion in the x & y directions are independent uy = u sin vxis constant = u cos x If not enough info in one dirn get time from the other dirn y Reference axes If time not involved Eki Ekf+ Ug ½mu2= ½mv2+ mgh u2= v2+ 2gh vy vx vx vx vx d vy vy t v u Ekmin = ½ mvx2 uy = u sin   = ux = ucos Speeds before hmax are the same as those after at each height v = u Eight points

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