Question

A student launches an elastic ball from the ground level directly toward a tall vertical wall. After the first launch, the ball bounces off the wall and then hits the ground. The student notices that the launch point is equidistant from the landing point and from the wall. The student then moves the launching point 20 m in the direction perpendicular to the wall and launches the ball again. After the ball bounces off the wall and lands, the student notices that the distance between the launch point and the landing point is the same as it was the first time. Finally, the student launches the ball away from the wall. How far apart are the launch point and the landing point this time? In all three cases, the initial velocity of the ball is the same.

Answer 1

Assuming the usual perfectly elastic collision, the horizontal component of the impact velocity is reversed. The ball lands exactly the same distance in front of the wall as it would have landed behind the wall. Let the distance from the first launch point to the wall be D. Since the ball hits the ground 2D in front of the wall, it would have hit the ground 2D behind the wall so that the horizontal range\ of the ball is 3D.

Now let us try moving the ball toward the wall. The distance from the launch point to the wall would be D – 20 m. The landing point would have to be 2D – 20 m from the wall, making the range 3D – 40 m, which is impossible. Therefore, the launch point must be moved 20 m away from the wall. Since the ball hits the ground a distance D from the launch point, the landing point is either 20 m or 2D + 20 m from the wall. The range would have to be either D + 40 m or 3D + 40 m. Since the range is 3D, only the first situation is possible and D = 20 m. Finally, if the ball is launched away from the wall, it will land 3D = 60 m away from the launch point.