At what projection angle will the range of a projectile equal its maximum height?
2. Relevant equations
vf2= vi2 + 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.
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.
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!