Projectile motion equation proof

[kubaanglin]

Homework Statement

Show that the launch angle θ is given by the expression:

θ=tan^-1(4h_max/R)

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

Homework Equations

h_max=v_i²sin²(θ)/2g
R=v_i²sin(2θ)/g

The Attempt at a Solution

I am trying to understand the significance of the ratio 4h_max/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?

 

[robphy]

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

 

[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?

 

[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.

 

[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: $R\sin(θ) = 4h_{max}\cos(θ)$ ?

 

[robphy]

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