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 (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
The distance between the two balls can be constant if and only if they have the same velocity. For the balls to have the same velocity, the velocity of the first ball released should clearly be same as the second ball when it is released at time (t) otherwise this scenario won’t be possible.
Let the initial velocity of the first ball be (u).
Let the second ball be released at time t = t seconds.
Then the initial velocity of the second ball and the final velocity of the first ball at time (t) is u/2
For the first ball
(v -> final velocity, u -> initial velocity, t->time).
(g is acceleration due to gravity)
The first ball would have gone up by 15 m till now because then only they will maintain a constant distance (s) of 15 m thereafter.
Taking (g) as 10m/s2, we get u = 20 m/s. Putting the value of (u) to get (t) which is equal to u/2g, we get t = 1 s.
Thus, the first ball was thrown at 20 m/s.
The second ball was thrown at 10m/s.
The time interval was 1 second.
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