A man is riding on a flatcar traveling at a constant speed of 9.10 (the figure )

Question

A man is riding on a flatcar traveling at a constant speed of 9.10 (the figure ). He wishes to throw a ball through a stationary hoop 4.90 above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of 10.8 with respect to himself.

When the ball leaves the man's hands, what is the direction of its velocity relative to the frame of reference of the flatcar?

When the ball leaves the man's hands, what is the direction of its velocity relative to the frame of reference of an observer standing on the ground?

Explanation / Answer

3.1: The height of the hoop is 4.90 m. The downwards acceleration due to gravity is 9.8 m/s^2. In order for the ball to travel horizontally through the hoop, it must achieve its maximum height at the hoop's level. I.e., its vertical component of velocity must be zero at the height of the hoop. We can use the formula v^2 = (v_0)^2 + 2ax ==> 0 = (v_0)^2 - 2gy ==> v_0 = sqrt(2gy) = sqrt(2*9.8*4.9) = 9.8 m/s, the initial vertical component of velocity. 3.2: This is the easiest part, now that we've solved 3.1. With the downwards acceleration due to gravity being 9.8 m/s^2, it clearly takes 1 s for the initial upwards velocity of 9.8 m/s^2 to be brought to a standstill, which we know is the condition for the ball to pass horizontally through the hoop. 3.3: If the vertical component of initial velocity is 9.8 m/s and the total velocity with respect to the thrower is 10.8 m/s, then the horizontal component with respect to the thrower must be sqrt(10.8^2 - 9.8^2) = 4.54 m/s. Then the total horizontal component of velocity is 4.54 + 9.10 = 13.64 m/s. Since the ball will only be in flight for 1 s before passing through the hoop, the ball must be released 13.64 m in front of the hoop.

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