The velocity at the maximum height of a projectile is half of its velocity of projection `u`. Its range on the horizontal plane is

Options

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

Answer

D

Explanation

Maximum height , `H_("max") = (u^2 sin^(2) theta)/(2g)`
Horizontal range , `R = (u^2 sin 2 theta)/g`
According to the given problem, `H_(max) = R/2`
`:. (u^2 sin^(2) theta)/(2g)=(u^2 sin 2 theta)/(2g)`
or `(u^2 sin^(2) theta)/(2g)=(u^2 sin 2 theta)/(2g)`
or ` tan theta = 2`
`:. sin theta =2/sqrt5 , cos theta =1/sqrt5`
`:. R =(2u^2 sin thetacos theta)/g=(2u^2 xx2/sqrt5xx1/sqrt5)/g=(4u^2)/(5g)`