# Finding the ideal angle for the maximum range of a projectile when elevated?

I was thinking about physics, and I thought of something. How do we find the ideal angle for the maximum range of a projectile when elevated?

Assumptions: Parabolic flight path No air friction

g = gravity constant, h = elevation height, v_i = velocity initial, l = launch angle


The time it takes to fall from h to 0 is defined as $$\frac{1}{2} * g * (t_0)^2 = h$$ $$\sqrt{\frac{2h}{g}} = t_0$$

The launch velocity for y can be defined as $v_i sin(l)$

The launch velocity for x can be defined as $v_i cos(l)$

The launch velocity for y will effect the range, find the maximum height of the object relative to its launch height $$v_i sin(l) * t_1 - \frac{1}{2} * g * (t_1)^2 = h_1$$

If we assume a parabolic arc, when relative 0 exists at t != 0 $$v_i sin(l) * t_1 - \frac{1}{2} g * (t_1)^2 = 0$$ $$v_i sin(l) * t_1 = \frac{1}{2} g * (t_1)^2$$ $$v_i sin(l) = \frac{1}{2} g t_1$$ $$t_1 = \frac{2 v_i sin(l)}{g}$$

Where the maximum height reached is at $\frac{t_1}{2}$ $$t_1 = \frac{v_isin(l)}{g}$$

The object is in the air for t_1 + t_0 seconds $$\frac{v_i sin(l)}{g} + \sqrt(\frac{2h}{g}) = t_1 + t_0 = t$$

The range can now be maximized $$range = v_i cos(l) * t$$

$$range = v_i cos(l) * \frac{v_i sin(l)}{g} + \sqrt{\frac{2h}{g}}$$

$$range = \frac{sin(l)cos(l)}{g} + \frac{cos(l)\sqrt{\frac{2h}{g}}}{v_i}$$

I am unsure how to continue to maximize this. I am also unsure if what I have done so far is correct.

EDIT: I think that I could theoretically remove all constant values so that my equation would become: (Am I allowed to do this)?

$$range = sin(l)cos(l) + cos(l)$$ $$range = cos(l)( sin(l) + 1 )$$

However, I'm not sure if I am able to do this to solve the problem.

• The range is not correct. The part until it reaches $h$ again should be $\frac{2v_i^2}{g}\cos l\sin l$. The time from $h$ to $0$ is not right either, for when it reaches height $h$ again the downward speed is not $0$, it is $v_i$. – André Nicolas Feb 27 '14 at 4:04

First, call its initial velocity $v$ and height $h$.

Here is the equation modelling the trajectory of the projectile: $y = h + x \tan \theta - \frac{g}{2v^2} x^2 (1 + \tan^2 \theta)$. We want to find where the projectile hits the ground, so let $y = 0$. We can now use calculus to find the maximum by differentiating w.r.t $\theta$.

$$x \sec^2 \theta + \tan \theta\frac{dx}{d \theta} - \left(\frac{gx}{v^2} \frac{dx}{d \theta} (1 + \tan^2 \theta) + \frac{gx^2}{2v^2} (2 \tan \theta \sec^2 \theta)\right) = 0$$

So $$\frac{dx}{d \theta} = \frac{x \sec^2 \theta \left(\frac{gx}{v^2} \tan \theta - 1\right)}{\tan \theta - \frac{gx}{v^2} (1 + \tan^2 \theta)}$$

So we find a critical value in $\tan \theta = \frac{v^2}{gx}$, and can substitute it back in the original equation:

$0 = h + \frac{v^2}{g} - \frac{g}{2v^2} x^2 - \frac{v^2}{2g}$ so $x = \frac{v}{g} \sqrt{v^2 + 2gh}$.

To find the angle, substitute in above to get $\theta = \tan^{-1} \left(\frac{v}{\sqrt{v^2 + 2gh}}\right)$

Notes: If you don't trust that this is a maximum, finding the second derivative is left as an exercise to the OP :)