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प्रश्न
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.
उत्तर
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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