# What is the relationship between launch height and the launch angle for maximum range?

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

• Hint: for a given elevation angle $\alpha$ and initial velocity $v_0$, the vertical component of the velocity is $v_{V0}=v_0\sin\alpha$ whilst the horizontal is $v_{H0}=v_0\cos\alpha$. The height changes in time as in vertical throw with initial velocity $v_{V0}$ and that determines the time – how long the projectile moves, The distance changes with a constant velocity $v_{H0}$, so the range is short if $\alpha$ is too small (the movement ends too soon) and if $\alpha$ is too big (the throw lasts long, but horizontal displacement changes too slow). Optimum is $\alpha=45^\circ$. – CiaPan Feb 16 '17 at 8:12
• I'm aware of the maximum range of a projectile at a constant height (ground level) , but I was wondering how changing the height may affect the optimum angle for the largest range. Thanks – connor Feb 16 '17 at 8:23
• I'm not sure what you mean. Are you considering e.g. shooting from a top of a building, so that the starting point and the landing plane are at different heights? – CiaPan Feb 16 '17 at 8:32
• Yeah that's what I mean. I want to know how changing the height, eg from the the top of a building to the ground, changes the angle which will give the greatest range. so from ground level the greatest range is given when the launch angle is 45 degrees but what angle gives the greatest range when changing the landing plane's – connor Feb 16 '17 at 8:46

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.

Assume projectile is launched from height of $h$. Let $v=$launch velocity, and $g=$gravity.

Trajectory equation for projectile is $$y-h=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$ Putting $t=\tan\theta$, and $k=\displaystyle\frac g{2v^2}$, equation becomes $$y-h=tx-k(t^2+1)x^2$$

Let $r=$horizontal range when projectile hits the ground, i.e. at $y=0$.

$$-h=tr-k(t^2+1)r^2\\ k(t^2+1)r^2-tr-h=0\qquad\qquad(1)$$ Differentiating w.r.t. t, $$k(t^2+1)2r\frac {dr}{dt}+k(2t)r^2-t\frac{dr}{dt}-r=0$$ At maximum $r$, $\displaystyle\frac{dr}{dt}=0$ $$r(1-2ktr)=0\\ \because r\neq0 \therefore r=\frac 1{2kt}$$ Substituting in (1) and rarranging gives $$t=\frac 1{\sqrt{1+4kh}}\\ \color{red}{\tan\theta^*=\frac 1{\sqrt{1+\frac {2gh}{v^2}}}}$$ This is the angle (or rather its tangent) which gives maximum range.

The maximum range is given by $$r^*=\frac {\sqrt{1+4kh}}{2k}\\ r^*=\frac {v^2}g\sqrt{1+\frac {2gh}{v^2}}=\color{orange}{\frac vg\sqrt{v^2+2gh}}$$

See illustration here.

The problem here is to find the optimal angle for the furthest range at a fixed launch velocity. Thisis the inverse of the problem in question $2660468$, which is the find the minimum velocity (and corresponding optimal angle) for a fixed range.
Also in the problem here the launch elevation is higher than the destination, while this is the opposition in question $2660468$. To make this equivalent we can simply assume that the projectile in question $2660468$ is launched in reverse, i.e. at the terminal velocity and angle.
Using the results in question $2660468$ (and using notation in this question), note that $$v^2=g(R-h)\\w^2=g(R+h)$$
Optimal angle $\phi$ is given by $$\tan\phi=\frac {R-h}k=\frac {R-h}{\sqrt{R^2-h^2}}=\frac {\frac {v^2}g}{\frac vg\sqrt{v^2+2gh}}=\color{red}{\frac v{\sqrt{v^2+2gh}}=\frac 1{\sqrt{1+\frac {2gh}{v^2}}}}$$ Maximum range $r^*$ is given by $$r^*=k=\frac {vw}g=\color{orange}{\frac {v{\sqrt{v^2+2gh}}}g}$$