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A projectile is thrown at angle beta wit...

A projectile is thrown at angle `beta` with vertical.It reaches a maximum height `H`.The time taken to reach the hightest point of its path is

A

`sqrt((H)/(g))`

B

`sqrt((2H)/(g))`

C

`sqrt((H)/(2g))`

D

`sqrt((2H)/(g cos beta))`

Text Solution

Verified by Experts

The correct Answer is:
B
`t=sqrt((2H)/(g))`
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