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Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec