A 2.7 kg ball is thrown upward with an initial speed of 20.0 m/s from the edge o

Question

A 2.7 kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball is thrown, a woman starts running away from the base of the cliff with a constant speed of 6.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored.

At what angle above the horizontal should the ball be thrown so that the runner will catch it just before it hits the ground, and how far does the woman run before she catches the ball?

Thanks to whoever helps out on this!

Explanation / Answer

Horizontal velocity of the ball = 20 cos A horizontal distance travelled by ball = 20cos A x t distance of the runner = 6 x t the runner will catch the ball at the same horizontal distance 20cos A x t = 6 x t cos A = 0.3 angle = 72.5deg (note: this seems too simple but hopefully correct)

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