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The trajectory of a projectile in a vert...

The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]`

Time of flight of the projectile is :-

A

`sqrt(3/10)s`

B

`sqrt(10/3)s`

C

`1s`

D

`2s`

Text Solution

Verified by Experts

The correct Answer is:
A
Time of fight `=(2u sin theta)/g=(2xxsqrt(10)xxsqrt(3)/2)/10=sqrt(3)/10`
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