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**Projectile Motion - V0 is 5 times v @ max height?!?**

Thanks prior, all...

## Homework Statement

A projectile's launch speed is 5 times its speed at maximum heigth. Find launch angle theta

_{0}.

## Homework Equations

Not sure which are relevant on this one, but the given for the section are:

x - x

_{0}= (v

_{0}cos theta

_{0})t,

y - y

_{0}= (v

_{0}sin theta

_{0})t - 1/2gt

^{2},

v

_{y}= v

_{0}sin theta

_{0}- gt,

v

_{y}

^{2}= (v

_{0}sin theta

_{0})

^{2}- 2g(y - y

_{0})

## The Attempt at a Solution

So, I've drawn the path, defining the maximum heigth as v

_{m}.

At v

_{m}, the vertical component, v

_{my}, must be 0.

v

_{mx}must be constant and would equal the same as it started,

v

_{mx}= v

_{ox}

In the problem, the initial velocity, v

_{0}is 5 times v

_{m}, so

v

_{0}= 5(v

_{m}) = 5(v

_{ox)}

Not sure where to go from here, I tried to break v

_{o}into its components, but just ended up with v

_{oy}

^{2}= 24v

_{ox}

^{2}

Any ideas?

Thanks again!