# Projectile motion equation proof

1. Oct 3, 2016

### kubaanglin

1. The problem statement, all variables and given/known data
Show that the launch angle θ is given by the expression:

θ=tan-1(4hmax/R)

where hmax is the maximum height in the trajectory and R is the range of the projectile.

2. Relevant equations
hmax=vi2sin2(θ)/2g
R=vi2sin(2θ)/g

3. The attempt at a solution
I am trying to understand the significance of the ratio 4hmax/R. I tried a couple of things, such as substitution and simplifying equations but have gotten nowhere. Can someone give me a tip on how to start this problem?

2. Oct 3, 2016

### robphy

Note that your expression for theta means tan(theta)=4hmax/R.
Note that tan(theta) is a slope.
Can you write tan(theta) in terms of hmax and R?

3. Oct 3, 2016

### kubaanglin

I plugged in the two equations and simplified:
$$\frac {\tan(θ)}{4} = \frac {h_{max}}{R} = \frac {\frac {v^2\sin^2(θ)}{2g}}{\frac {v^2\sin(2θ)}{g}} = \frac {\sin^2(θ)}{g\sin(2θ)} =$$
$$\frac {\sin^2(θ)}{2g\sin(θ)\cos(θ)} = \frac {\sin(θ)}{2g\cos(θ)} = \frac {\tan(θ)}{2g}$$
$$\frac {\tan(θ)}{4} \neq \frac {\tan(θ)}{2g}$$
What am I doing wrong? Are my initial equations incorrect?

4. Oct 3, 2016

### robphy

Recheck the third equal sign on the top row.

Usually, it is more elegant to not assume the equation that you are trying to obtain.
So, try to work it backwards by trying to first express tan(theta) in terms of sin(theta)/cos(theta), then see how one can get the expression you are trying to prove.

5. Oct 3, 2016

### kubaanglin

That was it! I made a careless error and treated $2g$ as $g^2$ in the third step.

Just to clarify, are you suggesting that I write proofs for these two equations: $\frac {h_{max}}{R} = \frac {\frac {v^2\sin^2(θ)}{2g}}{\frac {v^2\sin(2θ)}{g}}$ before using them?

If I am not going to assume the two equations or prove them, what do I do after this: $R\sin(θ) = 4h_{max}\cos(θ)$ ?

6. Oct 3, 2016

### robphy

Using the "relevant equations", eliminate vi and g... and somehow form tan(theta).