# Projectile at 1/4 height

1. Sep 16, 2011

### pereus

1. The problem statement, all variables and given/known data
The speed of a projectile when it reaches its maximum height is 1/4 the speed when the projectile is at 1/4 its maximum height.
What is the initial projection angle?
2. Relevant equations
v2 = v02 + 2a(delta s)
4Vx = v1/4
3. The attempt at a solution
I've tried substituting v1/4 for 4vx and some other things, but I lost those papers.

2. Sep 16, 2011

### PeterO

What is the significance of the speed at maximum height? What is the velocity at maximum height?

3. Sep 16, 2011

### pereus

I know that the velocity at the top is Vx. That is how I was able to make 4vx = to 4v1/4

4. Sep 16, 2011

### PeterO

What do you mean by Vx ? What is significant at the top?

5. Sep 16, 2011

### pereus

At the top the projectile is moving with the velocity Vx which is equal to v0cos(theta). There is no velocity in the y direction

6. Sep 16, 2011

### PeterO

OK.

You can actually analyse the second half of the projectile flight, as it mirrors the first [going down and going up work the same numerically] however.

At 1/4 the max height, there will be a vertical component to the velocity, and application of Pythagorus will enable you to find that, since the result is 4Vx.
You will get the vertical component in terms of Vx, but you are only using it to find an angle.
Once you have the vertical vel 1/4 way up - or if you like, 3/4 way down, you can find it at the beginning / end and again use trig to find the angle.

You could actually be using Potential Energy / Kinetic Energy / Total Energy to be working this out.