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A particle is projected vertically upwar...

A particle is projected vertically upwards from a point O on the ground. It takes time `t_(1)` to reach a point `A` at a height h above the ground, it continues to move and takes a time `t_(2)` to reach the ground. Find (a) h, (b) the maximum height reached and (c ) the velocity of the partical at the half of maximum height.

A

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

B

`"gt"_(1)t_(2)`

C

`(1)/(8)g(t_(1)+t_(2))^(2)`

D

`(1)/(2)"gt"_(1)t_(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`h = vt = (1)/(2) "gt"^(2)`
`rArr "gt"^(2) - 2vt +2h =0`
`t = (2v+-sqrt(4v^(2)-8gh))/(2g)`
`rArr t = (v+-sqrt(v^(2)-2gh))/(g)`
`t_(1) = (v-sqrt(v^(2) - 2gh))/(g) …(i)`
`t_(2) = (v+sqrt(v^(2)-2gh))/(g) …(2)`
From (1) & (2) `t_(1).t_(2) = (v^(2)-(v^(2)-2gh))/(g^(2))`
`rArr t_(1)t_(2) = (2h)/(g) rArr h =(1)/(2)"gt"_(1)t_(2)`
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