A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

Options

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

Answer

1

Explanation

`R=2H` given
Let `theta`=angle of projection
`(2v_(x)v_(y))/(g)=2.(v_(y)^(2))/(2g)`
`(v_(y))/(v_(x))=2`
`tantheta=2`
From tringle we can say that
`sintheta=(2)/(sqrt(5)),costheta=(1)/(sqrt(5))`
`:.` Range of projectile `R=(2v^(2)sintheta costheta)/(g)`
`=(2v^(2))/(g)xx(2)/(sqrt(5))xx(1)/(sqrt(5))=(4v^(2))/(5g)`