a juggler performs in a room whose ceiling is 9 ft above the levelof his hands h

Question

a juggler performs in a room whose ceiling is 9 ft above the levelof his hands he throws a ball vertically upward so that it just reaches theceiling
1. with what initial velocity does he throw the ball?
2. how many seconds are required for the ball to reach theceiling? he throws a second ball upward, with the same initialvelocity, at the instant the first ball touches the ceiling.
3. how long after the second ball is thrown do the two ballspass each other?
4. when the balls pass, how far are they above the jugglershands?
he throws a ball vertically upward so that it just reaches theceiling
1. with what initial velocity does he throw the ball?
2. how many seconds are required for the ball to reach theceiling? he throws a second ball upward, with the same initialvelocity, at the instant the first ball touches the ceiling.
3. how long after the second ball is thrown do the two ballspass each other?
4. when the balls pass, how far are they above the jugglershands?

Explanation / Answer

height of the ceiling h = 9 ft = 9 * 0.304 m = 2.736m (1). we know h = v ^ 2 / 2g from this initial velocity of the ball v =[ 2gh ] = 7.322 m / s (2). time t = v / g                 = 0.747 s (c).let after time t ' and height H both the ballspass each other distance travelled by ist ball in time t is S = (1/ 2) g t ' ^ 2 = 4.9 t ' ^ 2 distance travelled by 2 nd ball in time t ' is S ' =7.322 t ' - ( 1/ 2) g t ' ^ 2                                                                     =7.322 t ' -4.9 t '^ 2 we know h = S + S '          2.736 =4.9 t ' ^ 2 + 7.322 t ' - 4.9 f ' ^ 2         2.736 = 7.322 t'               t ' = 0.373 s

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