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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)]`

Maximum height H is :-

A

`8/3`

B

`3/8`

C

`sqrt(3)`

D

`2/sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Max. height `H=(u^(2) sin^(2) theta)/(2g),(10xx(sqrt(3)/2)^(2))/(2xx10)=3/8 m`
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