# Projectile Motion (find theta)

bengavinb20

## Homework Statement

The speed of a projectile when it reaches its maximum height is 0.31 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

## Homework Equations

h=(Vi^2 * sin^2 θ) / 2g

R=(Vi^2 * sin2θ) / g

## The Attempt at a Solution

I realize that i need to somehow come up with a ratio and set the two equal to each other. However, i'm having difficulty with the initial step. Maybe I could get a hint that could set me on the right track?

yands
Decompose the velocity vector.
At the top which component of velocity is zero?
Use this idea to find the ratio. As you mentioned

bengavinb20
So after decomposing the velocity vector I got:

Vi = -yf-at2 /t

Plugging that in to the equation for h...

h= ((-yf-at2)2*(sin2θ)) / 2g

and then

0.5h = 0.31((-yf-at2)2*(sin2θ)) / 2g

Am I on the right track?

yands
You see at the top point V is just Vx so can you relate this to the given information.

yands
I'm sorry for late response I got lectures :)
So you still solving the problem?