# Projectile Motion- What angle does range equal max height?

1. The problem statement, all variables and given/known data

At what projection angle will the range of a projectile equal its maximum height?

2. Relevant equations

Big 5

3. The attempt at a solution

So I attempted to solve this question by randomly subbing in equations into each other.

If i draw out a diagram to solve for the angle, I would need the x and y components.

Let V equal the total velocity.

viy= Vsinθ
Plugging into vfy2= viy2 + 2aydy , I get

vf2= V2sin2θ + 2aydy

Now I isolate for d, and end up with dy= (V2sin2θ)/2ay

Ok, after I got the displacement vertically, since I am trying to find the point where they both are at equal magnitude, I now solve for displacement horizontally.

Now using the range formula, I can determine the horizontal displacement.

dx= (V2sin2θ)/g

(g=ay)
With these two formulas, I substitute and…

(V2sin2θ)/2ay = (V2sin2θ)/ay

sin2θ = aysin2θ

sin2θ = ay(2sinθcosθ)

sinθ = 2aycosθ

sinθ/cosθ = 2ay

tanθ = 2(-9.8)

θ = tan-1(-19.6)

θ = 87°

However… this is not the answer that the answer key states. The answer shown is 76°, and I do not know where I have gone wrong, or if I am doing this right.

Sorry if it was a mess and confusing, thanks for the help!

http://ift.tt/MsOV7y