# Find the Angle of Projectile of the Horizontal Range :

If the horizontal range of a projectile is four times its maximum height, the angle of projection is

- 30
^{0} - 45
^{0} - Sin
^{-1}(1/4) - Tan
^{-1}(1/4)

**Answer: (2) 45 ^{o}**

**Explanation:-
**The standard equation,

R = u^{2} sin 2 θ/g

And h = u^{2} sin² θ/2g

The horizontal range of projectile is 4 times its maximum height.

Then, R = 4 h

u^{2}sin2θ/g = 4u^{2} sin^{2}θ/2g

sin2θ = 2 sin^{2}θ

2 sinθ cosθ = 2 sin^{2}θ (sin2θ = 2 sinθ cosθ )

cosθ = sinθ

Sinθ / cos θ = 1

tanθ = tan 45^{o}

θ = 45^{o}

Hence, the answer is (2) 45^{o}