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Projectile Motion Examples

Projectile Motion Examples. Example 3-6: Driving off a cliff!!. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m -high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are?.

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Projectile Motion Examples

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  1. Projectile Motion Examples

  2. Example 3-6: Driving off a cliff!! A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? y is positive upward, y0 = 0 at top. Also vy0 = 0 vx = vx0 = ? vy = -gt x = vx0t, y = - (½)gt2 Time to Bottom: t = √2y/(-g) = 3.19 s vx0 = (x/t) = 28.2 m/s

  3. Example 3-7: Kicked football A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. Calculate: a. Max height. b. Time when hits ground. c. Total distance traveled in the x direction. d. Velocity at top. e. Acceleration at top. θ0 = 37º, v0 = 20 m/s  vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

  4. Conceptual Example 3-8 v0x • Demonstration!!

  5. Conceptual Ex. 3-9: Wrong Strategy • “Shooting the Monkey”!! • Video Clip!!

  6. Example 3-10 • Range (R) of projectile  Maximum horizontal distance before returning to ground. Derive a formula for R.

  7. RangeR  the x where y = 0! • Use vx = vx0 , x = vx0 t , vy = vy0 - gt y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy • First, find the time t when y = 0 0 = vy0 t - (½)g t2  t = 0 (of course!) and t = (2vy0)/g • Put this t in the x formula: x = vx0 (2vy0)/g  R R = 2(vx0vy0)/g, vx0= v0cos(θ0), vy0= v0sin(θ0) R = (v0)2 [2 sin(θ0)cos(θ0)]/g R = (v0)2sin(2θ0)/g(by a trig identity)

  8. Example 3-11, A punt! • v0 = 20 m/s, θ0 = 37º • vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

  9. Proof that projectile path is a parabola • x = vx0 t , y = vy0 t – (½)g t2 Note: The same time t enters both equations!  Eliminate t to get y as a function of x. Solve x equation for t: t = x/vx0 Get: y = vy0 (x/vx0) – (½)g (x/vx0)2 Or: y = (vy0 /vx0)x - [(½)g/(vx0)2]x2 Of the form y = Ax – Bx2 A parabola in the x-y plane!!

  10. Ex. 3-12 a): Rescue Helicopter Drops Supplies A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), a) How far in advance of the recipients (horizontal distance) must the package be dropped?

  11. Ex. 3-12 b), c): Rescue Helicopter Throws Supplies A rescue helicopter wants to get a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. The helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), b) Someone in the helicopter throws the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? c) With what speed does the package land?

  12. That’s Quite an Arm! Problem: A stone is thrown from the top of a building at an angle θ0 =26° to the horizontal and with an initial speed v0 = 17.9 m/s, as in the figure. The height of the building is 45.0 m. a) How long is the stone "in flight"? b) What is the speed of the stone just before it strikes the ground?

  13. Example: The Long Jump Problem A long-jumper leaves the ground at angleθ0 = 20°above the horizontal at a speed ofv0 =8.0 m/s. a. How far does he jump in the horizontal direction? B. What is his maximum height?

  14. Stranded Explorers Problem:An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the picture. If the plane is traveling horizontally at v0 = 42.0 m/s at a height h = 106 m above the ground, where does the package strike the ground relative to the point at which it is released? v0 = 42 m/s h

  15. Problem 46

  16. Chapter 3, Problem 46 Solution Choose the origin at ground level, under where the projectile is launched, & up to be the positive ydirection. For the projectile: a. The time to reach the ground is found from Eq. 2-12b, with final height = 0. Choose positive time since the projectile was launched at time t = 0. b. The horizontal range is found from the horizontal motion at constant velocity.

  17. c. At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant The vertical component of velocity is found from Eq. 2-12a: d.The magnitude of the velocity is found from thexand y components calculated in part c. above.

  18. e. The direction of the velocity is so the object is moving f. The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from Eq. 2-12c.

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