# Projectile motion trajectories differing from 45 degrees

1. Apr 13, 2015

### bongobl

Hi guys, I am stuck with a problem here.
First, It is given that for 2-dimensional projectile motion, a trajectory of 45 degrees will yield the greatest range. However, how do I show that angles that differ from 45 degrees by the same amount will yield the same range? For example, the range of a 40 degree angle will equal that of a 50 degree angle?

I know that the range of a projectile as a function of time is given by V^2 * sin(2ø) / g where V is the initial velocity and ø is the angle. I just don't know how to prove that the angle 45 + c will give give the same range as 45 - c, can anyone help me please?

2. Apr 13, 2015

### nasu

3. Apr 13, 2015

### bongobl

Oh I see,
since sin(α + β) = sin(α)cos(β) + cos(α)sin(β),
and cos(90) = 0, I am left with sin(a)cos(B) in both cases,
thanks for the help!

4. Apr 14, 2015

### nasu

It is cos (c) in both cases. Sin of 90 is 1 and cos (-c)=cos (c).

5. Apr 18, 2015

### gracy

But how this answers @bongobl's question?