Question

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?

Answer 1

`~53.13` degrees

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 `v_0` at some angle `theta`.

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

`v_(y0) = v_0sin(theta)`

and the initial velocity in the x direction is given by

`v_(x0) = v_0cos(theta)`.

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:

`m (d^2x)/(dt^2) = 0`

Integrating gives

`(dx)/(dt) = C = v_(x0)`

`color(red) "Displacement in x as function of time"`

`implies x(t) = v_(x0)t = v_0 t cos(theta)`

Now look at the y direction:

`m(d^2y)/(dt^2) = -mg`

`(dy)/(dt) = -g t + C`

At `t=0`, `(dy)/(dt) = v_(y0) = v_0sin(theta)`

`implies (dy)/(dt) = -g t + v_0sin(theta)`

Integrate again

`color(red) "Displacement in y as function of time"`

`implies y(t) = -1/2 g t^2 + v_0 t sin(theta)`

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 `t = 0` which corresponds to before the projectile is launched and the other which shall be when it lands - the time of flight!

`0 = -1/2 g t^2 + v_0 t sin(theta)`

`1/2 g t^2 - v_0 t sin(theta) = 0`

`t(1/2 g t - v_0 sin(theta)) = 0`

For this to be true we must have either

`t = 0` or `( 1/2 g t - v_0 sin(theta) )= 0`

For time of flight (which we shall denote `t_f`) we take the second one:

`color(red) "Time of flight"`

`t_f = ( 2 v_0 sin(theta))/g`

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 `d`):

`color(red) "Range"`

`d = x(t_f) = v_0 cos(theta)*t_f = (2 v_0^2 sin(theta) cos(theta))/g = (v_0^2 sin(2 theta))/g`

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 `(dy)/(dt) = 0`

We shall denote the time at which the projectile is at max height as `t_m`, solving we get

`color(red) "Time at max height"`

`t_m = (v_0 sin(theta))/g`

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 (`h`) is simply a case of plugging in this value, ie

`h = y(t_m) = -1/2 g t_m^2 + v_0 sin(theta) t_m`

`h = -1/2 g ((v_0 sin(theta))/g)^2 + (v_0^2sin^2(theta))/g`

`color(red) "Max Height"`

`h = (v_0^2 sin^2(theta))/(2g)`

And now, finally, we are ready to answer the question. We want to know the value of `theta` such that `d = 3h`.

`therefore (v_0^2 sin(2theta))/g = (3 v_0^2 sin^2(theta))/(2g)`

`implies sin(2theta) = 3/2 sin^2(theta)`

`2sin(theta)cos(theta) = 3/2sin^2(theta)`

`sin(theta)(3/2sin(theta) - 2cos(theta)) = 0`

`sin(theta) = 0` or `tan(theta) = 4/3`

The first gives `theta = n pi` which doesn't make much sense so we look at the second:

`theta = arctan(4/3) ~= 53.13` degrees (there are other solutions to this but this is the first and only one that makes physical sense)