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 10 m/s at 33° above the horizontal. A friend of yours, standing on the court 13 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?

Explanation / Answer

Horizontal velocity of the ball 10 cos 33 = 6.949 m/s
Vertical velocity of the ball 10 sin 33 = 3.964 m/s

Time to fall through a vertical distance of (13-6) m is found using

S = ut + ½ at²

- 7 = 3.964*t – 4.9 t²
t = 1.583 s

Horizontal distance traveled in this time is 6.949*1.583 = 11m

Since the friend of yours, standing on the court 11 m from the point directly beneath you, she need not move from that point.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.