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?
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:
Now look at the y direction:
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