# Initial velocity of projectile given angle and max height

## Homework Statement

A projectile was fired across level ground at an initial angle of 43.00 degrees above the horizontal. During its flight, it reached a maximum height of 2974 metres. What was the initial speed of this projectile?

## Homework Equations

I assume that the correct equation to use is Vy2 = V0y² - 2g(y - y0).

## The Attempt at a Solution

To be honest I'm not even sure how to go about it. Given the equation, I would rearrange for V0y² however that doesn't seem to work at all. I'm not even sure what the value of Vy has to be. I also suspect the angle comes into play, but I'm not sure what it can tell me. Any pointers greatly appreciated.

At the top of the flight, there is no vertical component of velocity, so $v_y = 0$ there. You are also correct that the angle is involved. Indeed, $v_o\sinθ= v_{oy}$ and it is this $v_o$ that you are solving for.