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A cricket ball thrown across a field is ...

A cricket ball thrown across a field is at heights `h_(1)` and `h_(2)` from the point of projection at time `t_(1)` and `t_(2)` respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

A

`(h_(1)r_(2)^(2) - h_(2)r_(1)^(2))/(h_(1)t_(2) -h_(2)t_(1))`

B

`(h_(1)t_(1)^(2) + h_(2)t_(2)^(2))/(h_(2)t_(1) + h_(1)t_(2))`

C

`(h_(1)t_(2)^(2) + h_(2)t_(1)^(2))/(h_(1)t_(2) + h_(2)t_(1))`

D

`(h_(1)t_(1)^(2) - h_(2)t_(2)^(2))/(h_(1)t_(1) - h_(2)t_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
Let a cricket ball be thrown in upward direction with velocities u at an angle `theta` with the horizontal.
As per question,
`h_(1) = u sin theta t_(1) - 1/2 "gt"_(1)^(2)`
or `u sin theta t_(1) -h_(1) + 1/2 "gt"_(1)^(2)`………(i)
and `h_(2) = usin theta t_(2) - 1/2 "gt"_(2)^(2)`........... (ii)
Divide eqn. (i) by equn. (ii), we get
`t_(1)/t_(2) = (h_(1) + 1/2 "gt"_(1)^(2))/(h_(2) + 1/2 "gt"_(2)^(2)) h_(2)t_(1) + 1/2 "gt"_(2)^(2) t_(1) = h_(1)t_(2) + 1/2"gt"_(1)^(2)t_(2)`............. (iii)
Time of flight, `T = (2u sin theta)/g`
`T = 2/g [(h_(1) + 1/2"gt"_(1)^(2))/t_(1)]` using (i)
`2/g h_(1)/t_(1) + t_(1) = h_(1)/t_(1) (t_(1)t_(2)^(2) - t_(1)^(2) t_(2))/(h_(1)t_(2) - h_(2)t_(1)) + t_(1)` (using (iii))
`=(h_(1)t_(2)^(2) - h_(2)t_(1)^(2))/(h_(1)t_(2) - h_(2)t_(1))`
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