A juggler performs in a room whose ceiling is 3.0 m above thelevel of his hands.

Question

A juggler performs in a room whose ceiling is 3.0 m above thelevel of his hands. He throws a ball upward so that it just reachesthe ceiling. a) What is the initial velocity of the ball? b) Whatis the time required for the ball to reach the ceiling? At theinstant when the first ball is at the ceiling, he throws a secondball upward with two-thirds the initial velocity of the first. c)How long after the second ball is thrown did the two balls passeach other? d) At what distance above the juggler's hand do theypass each other?

Explanation / Answer

a) You know : V = 0 when d =3.0m .. and the acceleration is -g (-9.8m/s²) for upward vertical movement. (v² = U² + ...) b) Try d = 3 = average velocity x time av. vel. = (V + U) /2 .... (V=0) c) Balls pass when ball falling has travelled (d1) = ½g t² down and rising ball has travelled (d2) = (2/3U)t - ½gt² d1 + d2 = 3m ..........(½gt²) + (2/3Ut - ½gt²) = 3 ...........2/3Ut =3 ......... hence find t d) put value for t into equation for d2 above ... = distance above juggler's head. U = 7.67 .. t = 0.78 .. c) t = 0.58 .. d2 = 1.32

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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