A gun is firing bullets with velocity `v_0` by rotating it through `360^@` in the horizontal plane. The maximum area covered by the bullets is

Options

A. `pi((v_0^(4))/g^(2))`
B. `pi((v_0^(2))/(2g))^(2)`
C. `pi(v_0/g)^(2)`
D. `pi(v_0/(2g))^(2)`

Answer

A

Explanation

`R=(u^(2)sin 2 theta)/g`
Range R is maximum when ` sin 2 theta=1`
`R_(max)=(u^(2))/g`
Maximum area coverted `=pi(R_(max))^(2)=pi((u^(2))/g)^(2)`