A projectile is thrown with an initial velocity of `(a hati +b hatj) ms^(-1)`. If the range of the projectile is twice the maximum height reached by it, then

Options

A. `b = a/2`
B. `b = a`
C. `b = 2a `
D. `b = 4a`

Answer

C

Explanation

Given initial velocity , `vecu = a hati + b hatj `
`:. u_(x) = a, u_(y) = b`
Time of fight, `T = (2u_y)/(g) =(2b)/g`
Maximum height , H `=(u_y^2)/(2g) = (b^2)/(2g)`
Range , `R=(2u_xu_y)/g=(2ab)/g`
`:. R = 2H ` (Given )
`:. (2ab)/g = 2xx (b^2)/(2g) " or " 2a = b `