# Projection angle using max. height w/ proportions only

Figures that the only problem I have trouble with is the one the book considers to be "easy":

## Homework Statement

The speed of a projectile when it reaches its maximum height is one half 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 theta)/2g

## The Attempt at a Solution

Complete brain fart--I don't even know where to start, beyond trying to divide h/2, which would be (vi^2 sin^2 theta)/4g, correct? That's about as far as I got. I just keep miring myself the further into it I go.

## Answers and Replies

Kurdt
Staff Emeritus
Science Advisor
Gold Member
Speed of course is the magnitude of velocity. At its maximum height a projectile only has one component of velocity. Is that enough to get you started?

Try breaking velocity into x and y components @ 1/2 max height and then finding speed from there the "sqrt of the sum of the squares".