Question

what is the angle of projection if horizontal range is 4 times the maximim height attained..???

Answer 1

See the answer below in attachment

Answer 2

So you have the formulas as 

for Range( R ) = $\frac{u^2sin2\theta}{g}$

Max. Height (H) = $\frac{u^2sin^2\theta}{2g}$

u = initial velocity 
g = acceleration due to gravity
θ = angle of projection

so according to question R = 4H

so $\frac{u^2*sin2\theta}{g}$ = 4$\frac{u^2sin^2\theta}{2g}$

⇒$\frac{4}{4}=tan\theta$

θ = 45° ANSWER