A body is thrown with some velocity from the ground. Maximum height when it is thrown at `60^(@)` to horizontal is 90m. What is the height reached when it is thrown at `30^(@)` to the horizontal:-

Options

A. 90m
B. 45m
C. 30m/s
D. 15m

Answer

C

Explanation

`h=(u^(2)sin^(2)theta)/(2g)rArrh_(1)=90=((u sin 60^(@))^(2))/(2g)`
`90=(u^(2))/(2g)xx[sqrt(3)/(2)]^(2)rArr(u^(2))/(2g)=120`
`h_(2)=((u sin 30^(@))^(2))/(2g)=120xx(1)/(4)=30m`