- #1

Acnhduy

- 31

- 1

## Homework Statement

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

## Homework Equations

Big 5

v

_{f}

^{2}= v

_{i}

^{2}+ 2ad

## 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!