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Projectile Motion Examples

Projectile Motion Examples. Example 4.3: The Long Jump. Problem: A long-jumper (Fig. 4.12) leaves the ground at an angle θ i = 20 ° above the horizontal at a speed of v i = 8.0 m/s . a) How far does he jump in the horizontal direction?

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Projectile Motion Examples

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  1. Projectile Motion Examples

  2. Example 4.3: The Long Jump Problem:A long-jumper (Fig. 4.12) leaves the ground at an angle θi = 20° above the horizontal at a speed of vi =8.0 m/s.a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached?

  3. Example: Driving off a cliff!! y is positive upward, yi = 0 at top. Also vyi = 0 • How fast must the motorcycle leave the cliff to land at xf = 90 m, yf = -50 m? vxi = ? vx = vxi = ? vy = -gt x = vxit, y = - (½)gt2 Time to Bottom: t = √2yf/(-g) = 3.19 s vxi = (xf/t) = 28.2 m/s

  4. Kicked football θi = 37º, vi = 20 m/s  vxi = vicos(θi) = 16 m/s, vyi= visin(θi) = 12 m/s a. Max height? b. Time when hits ground? c. Total distance traveled in the x direction? d. Velocity at top? e. Acceleration at top? vf vyi vxi

  5. Conceptual Example vyi vxi • Demonstration!! vyi vi vxi

  6. Conceptual Example: Wrong Strategy vi  • “Shooting the Monkey”!! • Demonstration!!

  7. Example xi = 0 yi = 0 θi • Range (R) of projectile  Maximum horizontal distance before returning to ground. Derive a formula for R. θi θi1 θi1 θi2

  8. RangeR  the x where y = 0! • Use vxf = vxi , xf = vxi t , vyf = vyi - gt yf = vyi t – (½)g t2, (vyf) 2 = (vyi)2 - 2gyf • First, find the time t when y = 0 0 = vyi t - (½)g t2  t = 0 (of course!) and t = (2vyi)/g • Put this t in the x formula: xf = vxi (2vyi)/g  R R = 2(vxivyi)/g, vxi= vicos(θi), vyi= visin(θi) R = (vi)2 [2 sin(θi)cos(θi)]/g R = (vi)2sin(2θi)/g(by a trig identity)

  9. Example 4.5: That’s Quite an Arm! Problem: A stone is thrown from the top of a building at an angle θi =26° to the horizontal and with an initial speed vi = 17.9 m/s, as in Fig. 4.14. 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?

  10. Example: A punt! • vi = 20 m/s, θi = 37º • vxi = vicos(θi) = 16 m/s, vyi= visin(θi) = 12 m/s

  11. Proof that projectile path is a parabola • xf = vxi t , yf = vyi 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 = xf/vxi Get: yf = vyi (xf/vxi) – (½)g (xf/vxi)2 Or: yf = (vyi /vxi)xf - [(½)g/(vxi)2](xf)2 Of the form yf = Axf – B(xf)2 A parabola in the x-y plane!!

  12. Problem vi = 65 m/s 65

  13. Example 4.6: The 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 vi = 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? vi = 65 m/s h

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