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A ball is thrown upward at an unknown angle 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 catches it just before it hits the ground, and how far does the woman run before she catches the ball?

Respuesta :

Answer:

distance by the runner = 6 t

θ = 72.54°

Explanation:

initial velocity of ball = 20 m/s

speed of the runner = 6 m/s

horizontal velocity in the projectile = 20 cos θ

horizontal distance traveled by the ball =  20 cos θ × t

t is the time at which the velocity is to be calculated

distance by the runner = 6 × t

to catch the ball distance by the ball should be equal to the distance traveled by the runner.

20 cos θ × t = 6 × t

cos θ =  0.3

θ = 72.54°

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