A stone is thrown vertically at a speed of `30 ms^(-1)` making an angle of `45^(@)` with the horizontal. What is the maximum height reached by the stone ? Take `g=10 ms^(-2)`.

Options

A. 15 m
B. 30 m
C. 10 m
D. 22.5 m

Answer

D

Explanation

Maximum height reached by the stone is
`H_("max")=(u^(2)sin^(2)theta)/(2g)`
Here, `u=30 ms^(-1), theta=45^(@), g=10ms^(-2)`
`:.H_("max")=((30)^(2)sin^(2)45^(@))/(2xx10)=(30xx30xx(1/sqrt(2))^(2))/(2xx10)`
`=(45)/(2)=22.5m`