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Question
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.
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Solution
Let the speeds of the two balls (1 and 2) be v1 and v2 where
If v1 = 2v, v2 = v
If y1 and y2 and the distance covered by balls 1 and 2, respectively before coming to rest then
`y_1 = v_1^2/(2g) = (4v^2)/(2g)` and `y_2 = (v_2^2)/(2g) = v^2/(2g)`
Since, `y_1 - y_2 = 15 m (4v^2)/(2g) - v^2/(2g)` = 15 m or `(3v^2)/(2g)` = 15 m
or `v^2 = sqrt(5m xx (2 xx 10))` m/s2
or v = 10 m/s
Clearly, v1 = 20 m/s and v2 = 10 m/s
As `y_1 = v_1^2/(2g) = (20 m)^2/(2 xx 10 m 15)` = 20 m
`y_2 - y_1 - 15` m = 5 m
It t2 is the time taken by the ball 2 toner a distance of 5 m, then from `y_2 = v_2^t - 1/2 "gt"_2^2`
5 = `10t_2 - 5t_2^2` or `t_2^2 - 2t_2 + 1` = 0
Where t2 = 15
Since t1 (time taken by ball 1 to cover the distance of 20 m) is 2s, the time interval between the two throws
= t1 – t2
= 2s – 1s
= 1s
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