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

Question

A juggler performs in a room whose ceiling is 3.0m above the level of his hands. He throws ball upward so that it just reaches the ceiling.
1) What is init velocity of ball?
2) What is time required for ball to reach ceiling?

At the instant when the first ball is at the ceiling, juggler throws a second ball upward with 2/3 (two-thirds) the init velocity of the first.
3) How long after the second ball is thrown did the two balls pass each other?
4) At what distance above the juggler's hand do they pass each other?

Explanation / Answer

1) You know : V = 0 when d =3.0m .. and the acceleration is -g (-9.8m/s²) for upward vertical movement. (v² = U² + ...) 2) Try d = 3 = average velocity x time av. vel. = (V + U) /2 .... (V=0) 3) 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 4) 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

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