A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is :

Options

A. `(4v^(2))/(5g)`
B. `(4g)/(5v^(2))`
C. `(4v^(3))/(5g^(2))`
D. `(4v)/(5g^(2))`

Answer

A

Explanation

`H=(v^(2) sin^(2) theta)/(2g) and R=(v^(2) sin 2 theta)/g`
Since `R=2H, (v^(2) sin 2 theta)/g=2xx(v^(2) sin^(2) theta)/(2g)`
or `2 sin theta cos theta= sin^(2) theta or tan theta=2`
`R=v^(2)xx2/gxxsin theta cos theta`
`=(2v^(2))/g =2/(sqrtg)xx1/(sqrt5)=(4v^(2))/(5g)`