My physics teacher gave this question to our class and it has me stumped. I cant find any written examples on the internet talking about the relationship between an elevated projectile and the optimum angle for the furthest range. was hoping someone might be able to help me out. Thanks
[Math] the relationship between launch height and the launch angle for maximum range
physicsprojectile motion
Related Solutions
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 :)
There seems to be a typo in the problem : The $V = 2\sqrt{gd}$ and then the rest follows as given below in the solution. I think I got what the book answer is.
Solution as follows:
Best Answer
Have you tried to search the net for 'projectile initial height'?
A few first hits contain e.g.
which all seem to give answer to your problem.
EDIT
Also Maximum range of a projectile (launched from an elevation) here, at Math SE, seems to be the same problem as yours.