A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is

Options

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

Explanation

Greatest height attained `h=(u^(2) sin^(2) 0)/(2g)…….(1)`
Horizontal range `R=(u^(2) sin20)/(g) =(2u^(2) sin 0 cos 0)/(g) ……(2)`
Given that R=2h
`rarr (2u^(2) sin 0 cos 0)/(g) =(2u^(2) sin^(2)0)/(2g) rarr tan 0=2" …….(3)`
Hence `sin 0=2//sqrt(5), cos 0=1//sqrt(5)`,
From (2) `R=4u^(2)//5g`