A person is riding on a flatcar traveling at a constant speed v1= 15 m/s with respect to the ground. He wishes to throw a ball through a stationary hoop in such a manner that the ball will move horizontally as it passes through the hoop.

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Cool, I hope it works out for him.

Imagine the person throws the ball straight up in the air--he does not throw it forward or backward, just straight up--this is the vertical direction. Imagine also that the hoop is exactly 10m above the car. To figure out where he should be when he throws the ball we have to remember that the horizontal component of the ball's velocity is NOT EFFECTED by the throw. The ball is moving horizontally along with the car at 15m/s. This does not change. Similarly, the vertical motion of the ball is not effected by the motion of the car. If we imagine there is no air resistance, then the ball's vertical motion (up/down direction) is only effected by the initial throw and the acceleration of gravity towards the ground. How long will it take the ball to reach the height of the hoop, 10 m above the ground? When the ball reaches its apex--the highest point it will reach--its vertical velocity is momentarily zero, so let vf, final velocity, equal zero. You know the distance (10m), the acceleration (-9.8), and the final velocity (0). Using the equations in your book, you can solve for the initial velocity that he should use to throw the ball. Then you can solve for the amount of time it takes the ball to reach its peak. Now that you know the time it takes the ball to reach its peak, what do you do? Well, the ball moves HORIZONTALLY at 15m/s. And it takes some amount of time (solved for above) to reach its peak. since distance=rate x time, you can solve for the distance he should be away from the point on the ground just below the hoop when he throws it. If you need the equations, write back.

The ball moves in projectile motion. When it is moving horizontally, v(y) =0 +x is directed to the right, +y is directed upward, a(x)= 0, a(y) =-g v₀(y)=sqrt(2gh) = sqrt(2•9.8•4) = 8.85 m/s v= v₀(y)-gt At the top point v=0 => t= v₀(y)/g =8.85/9.8 =0.9 s. The horizontal component of the ball’s velocity relative to the man is sqrt{v₂²-v₀(y) ²} = sqrt{18²-8.85² } =15.67 m/s: the horizontal component of the ball’s velocity relative to the hoop is 15.67 +15 =30.67 m/s, and the man must be 30.67•0.9 = 27.6 m in front of the hoop at release. Relative to the flat car, the ball is projected at an angle tanα = v₀(y)/15.67 =8.85/15.67=0.565 α=29.46⁰ Relative to the ground the angle is tanβ =8.85/(15.67 +15) =8.85/30.67= 0.289 β=16.1⁰ In both frames of reference the ball moves in a parabolic path. The only difference between the description of the motion in the two frames is the horizontal component of the ball’s velocity.

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