# How to calculate optimal angle for maximum range of a projectile - on a hill

Basically, the following problem was tasked to the first-time physics students:

Calculate the angle needed for the maximum range of a projectile, except in this case the projectile is on a hilltop and there is a steady angle going downwards at 8 degrees.

I haven't found a sufficient way to figure this out; I'm not sure if calculus is needed?

What are the steps I should take to solve this problem? Not sure how to find the relationship between the extra distance because of the angle, and how to maximize it other than guess and check using the kinematic equations and the equation given for range. Any help is greatly appreciated!

• Break the motion into the horizontal and vertical components. The solve for when the $(x,y)$ coordinate is on the hill surface. Oct 6, 2019 at 22:12
• This is equivalent to being on level ground with the constant gravitational force directed somewhere other than straight down.
– amd
Oct 6, 2019 at 23:42

Use the horizontal and vertical distance equations,

$$x=v_0t\cos\theta,\>\>\>\>\>\>y=v_0t\sin\theta -\frac12 gt^2$$

where $$\theta$$ is the launching angle with respect to the horizon and $$v_0$$ is the initial speed. Eliminate the time variable to obtain,

$$y=\tan\theta\> x-\frac{gx^2}{2v_0^2\cos^2\theta}$$

Due to the hill of angle $$\alpha$$, the relationship $$y_L=-x_L\tan\alpha$$ holds at the landing point $$(x_L,y_L)$$. Plug it into the above equation to obtain the equation for $$x_L$$,

$$-\tan\alpha \>x_L=\tan\theta\> x_L-\frac{gx_L^2}{2v_0^2\cos^2\theta}$$

which leads to the solution

$$x_L = \frac{2v_0^2}{g}(\sin\theta\cos\theta + \tan\alpha\cos^2\theta)$$

For maximal range, take the derivative of $$x_L$$ with respect to $$\theta$$ and set it to zero,

$$\cos 2\theta_m - \tan\alpha\sin 2\theta_m=0$$

Solve for the optimal angle $$\theta_m$$,

$$\theta_m = \frac 12 (90^\circ - \alpha)$$

For the specified hill slope angle $$\alpha = 8^\circ$$ in the post, the optimal launching angle is $$41^\circ$$.