# 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

v_{f}^{2}= v_{i}^{2} + 2ad

**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.

v_{iy}= Vsinθ

Plugging into v_{fy}^{2}= v_{iy}^{2} + 2a_{y}d_{y} , I get

v_{f}^{2}= V^{2}sin^{2}θ + 2a_{y}d_{y}

Now I isolate for d, and end up with d_{y}= (V^{2}sin^{2}θ)/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.

d_{x}= (V^{2}sin2θ)/g

(g=a_{y})

With these two formulas, I substitute and…

(V^{2}sin^{2}θ)/2a_{y} = (V^{2}sin2θ)/a_{y}

sin^{2}θ = a_{y}sin2θ

sin^{2}θ = a_{y}(2sinθcosθ)

sinθ = 2a_{y}cosθ

sinθ/cosθ = 2a_{y}

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

## Leave a comment