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A particle is projected vertically upwar...

A particle is projected vertically upwards in vacuum with a speed `u`.

A

When it rises to half its maximum height, its speed becomes `u//2`.

B

When it rises to half its maximum height, its speed becomes `u//sqrt(2)`.

C

The time taken to rise to half its maximum height is half the time taken to reach its maximum height.

D

The time taken to rise to three`-` fourth of its maximum height is half the time taken to reach its maximum height.

Text Solution

Verified by Experts

The correct Answer is:
B, D
At maximum height, velocity `=0`
`H=(u^(2))/(2g)&`
At height `h=H//2, V^(2)=u^(2)-2gh`
`V^(2)=u^(2)-2g.(u^(2))/(4g), V^(2)=(u^(2))/(2)rArrV=(u)/(sqrt(2))`
Time taken to rise to maximum height `T=(u)/(g)` for height `h=(H)/(2)t=((u-u//sqrt(2)))/(g)=((sqrt(2)-1)u)/(sqrt(2)g)`
Time taken to rise to `(3)/(4)H=T-` time taken to fall down by `(H)/(4)`
`=T-(T)/(2)=(T)/(2)`
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