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A ball is thrown up, it reaches a maximu...

A ball is thrown up, it reaches a maximum height and then comes down. If `t_(1) and t_(2)(t_(2)gt t_(1))` are the times that the ball takes to be at a particular height, then the time taken by the ball to reach the highest point is

A

`t_(1)+t_(2)`

B

`t_(2)-t_(1)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
`v=u-g t_(1)` . . (1)
`-v=u-g t_(2)` . .. (2)
(1) and (2) ading we get
`0=2u-g(t_(1)+t_(2))`
`u=(g)/(2)(t_(1)+t_(2))` . . . (3)
Generally v=u+at
At highest point v=0, a=-g
From eq. (3) we get
`therefore 0=(g)/(2)(t_(1)+t_(2))- g t`
`therefore t=((t_(1)+t_(2)))/(2)`.
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