The kinetic energy of a projectile at its highest position is `K`. If the range of the projectile is four times the height of the projectile, then the initial kinetic energy of the projectile is .

Options

A. `sqrt2K`
B. `2 K`
C. `4 K`
D. `2sqrt2 K`

Answer

B

Explanation

At highest point,
`(1)/(2)m u_(x)^(2)=K`
`:. u_(x)=sqrt((2K)/m)`
`R=4 H`
`(2u_(x)u_(y))/g=(4u_(y)^(2))/(2g)`
`:. u_(y)=u_(x) =sqrt((2K)/m)`
Now, `K_(i)1/2m u^(2) =1/2m(u_(x)^(2)+u_(y)^(2))`
`=1/2m((2K)/m+(2K)/m)`
`=2K`.