A body is projected at an angle `theta` so that its range is maximum.If `T` is the time of flight then the value of maximum range is (acceleration due to gravity =`g`)

Options

A. `(g^(2)T)/2`
B. `(gT)/2`
C. `(gT^(2))/2`
D. `(g^(2)T^(2))/2`

Answer

C

Explanation

for maximum range `theta=45^(@)`
`R_(max)=u^(2)/g` when`theta=45^(@),T=(2usintheta)/g`