# A projectile is fired in such a way that it's horizontal range is equal to three times it's maximum height. What is the angle of projection?

##### 1 Answer

#### Explanation:

I've done full derivations here. If you're not interested I've put coloured titles above the important results that I've used.

So the projectile is fired with some initial velocity

This means that the initial velocity in the y direction is given by

and the initial velocity in the x direction is given by

We now use Newton's 2nd law in the two independent directions. Define upwards as the positive y direction and motion to the right as the positive x direction. Let's consider the x direction first:

Integrating gives

Now look at the y direction:

At

Integrate again

Now we have basic equations for the position in each direction as functions of time. From this we need to obtain expressions for the range and the maximum height in order to solve the problem.

To find the range, we need to determine the time of flight. We do this by setting y = 0. This will give us two times, one at

For this to be true we must have either

For time of flight (which we shall denote

Now that we have an expression for the time of flight, we can sub it into our formula for x distance to find the range (denoted

Almost there, we just need to find an expression for the maximum height. We do this by finding the critical point of the y value, ie we set

We shall denote the time at which the projectile is at max height as

This is exactly half the time taken for the projectile to reach the end (can you see why this makes sense?).

Finding the max height (

And now, finally, we are ready to answer the question. We want to know the value of

The first gives