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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)]`

Angle of projection `theta` is :-

A

`30^(@)`

B

`60^(@)`

C

`45^(@)`

D

`sqrt(3)` rad

Text Solution

Verified by Experts

The correct Answer is:
B
`y=sqrt(3)x-2x^(2)`
Trajectory equation is `y=x tan theta-(gx^(2))/(2u^(2) cos^(2) theta)`
`tan theta=sqrt(3)rArr theta=60^(@)` & `(g)/(2u^(2)cos^(2) theta)=2`
`rArr u=5/(2xx1/4)=sqrt(10)`
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