A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity
of half the first. The vertical gap between first and second ball is +15m at t = 2s. The gap is found to remain constant. Calculate the velocity with which the balls
were thrown and the exact time interval between their throw.
Let the speed of ball 1 = u1 = 2u m/s
Then the speed of ball 2 = u2 = u m/s
The height covered by ball 1 before coming to rest = h1
The height covered by ball 2 before coming to rest = h2
We know that,
v2 = u2 + 2gh
v2 = 2gh
h = v2/2g
Therefore, h1 = 4u2/2g
h2 = u2/2g
From question, h1 –h2 = 15 m
Therefore,
4u2/2g – u2/2g = 15
u = 10 m/s
h1 = 20 m
h2 = 5 m
And time taken is,
t1 = 2 sec
t2 = 1 sec