You throw a ball from the balcony onto the court in the basketball arena. You re

Question

You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 6 m above the court, with an initial velocity equal to 8 m/s at 32° above the horizontal. A friend of yours, standing on the court 8 m from the point directly beneath you, waits for a period of time after you release the ball and then begins to move directly away from you at an acceleration of 3 m/s2. (She can only do this for a short period of time!) If you throw the ball in a line with her, how long after you release the ball should she wait to start running directly away from you so that she'll catch the ball exactly 1 m above the floor of the court? PLEASE SHOW WORK and explain

Explanation / Answer

Horizontal velocity of the ball 8 cos 32 = 6.784 m/s
Vertical velocity of the ball 8 sin 32 = 4.23 m/s

Time to fall through a vertical distance of (6-1) m is found using

S = ut + ½ at²

- 5 = 4.23*t – 1.5 t²
t = 3.71 s

Horizontal distance traveled in this time is 6.78*3.71 = 25.15m

Since the friend of yours, standing on the court 8 m from the point directly beneath you, she need not move from that point.

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