**Below is a proof, written by Kian Mayne, on how the optimum angle to throw/ kick or shoot something to make it travel the furthest distance horizontally possible is 45 degrees!**

Applying the equations of linear motion (SUVAT) for the horizontal (we want to maximise the distance travelled along the x-axis):

– hmm…it’s going to be flying from when it’s kicked/thrown to when it hits the ground

– neglecting air resistance

– resolving for the x-axis

– don’t know and can’t find out

– what we want to find out

Looks like we need to find ! As mentioned above – we can find this from how long it stays in the air. The easiest way to do this is to double the time taken to reach the highest point. More SUVAT (but for the vertical y-axis!):

– what we want to find out

– we’re taking upwards as positive

– resolving for the y-axis

– velocity is zero at the apex of the throw/kick/whatever

– unknown and irrelevant

We have all the variables except therefore we need the not-s equation:

Excellent! We now have a constant expression for that we can use in the horizontal SUVAT equation (for which we use the not-v equation):

Substituting from earlier:

Awesome! We now have an equation for in terms of constants and . Finding the maximum can be done in two ways – one simple and one cooler and more mathematically rigorous.

The simple way is as follows:

The part is a constant and thus can be ignored.

Therefore to maximise the function, we need to maximise

You can find the maximum by drawing the function (below)

As you can see, the maximum occurs at 45°.

Boorrring. Let’s find the maximum via differentiation! When differentiating, we find a maximum by using the fact that, at a maximum or minimum (or point of inflection), the gradient (found by differentiating) is 0. OK, let’s see:

As is constant, we can factor it out:

Using the product rule:

Cool – now let’s set it equal to 0:

As is a constant, we can safely divide through by it (there are no solutions for in it):

Now using the following trigonometric identity:

Now substituting:

Square root both sides (we can ignore the negative root as that is in the opposite direction):

of both sides:

There! FYI We can safely solve for as anything higher or lower would be the opposite direction.

**Thanks Kian!**