Cannon on a Cliff

45 degrees is optimum on flat ground because it gives the right balance of lift and forward thrust. If you start higher up, you get some lift for free so can devote more of your charge to forward thrust and can aim lower than 45 degrees.

(If you start in a hole you need more lift and a higher angle.)

Projectile motion is parabolic with a decreasing slope.

When the cannon and ground are at equal heights, the optimal angle of launch is 45 degrees but it is better to think that the reason for this is that the angle of impact is 45 degrees. This is the angle after which the motion is predominantly vertical, wasting effort if the goal is to maximise horizontal distance.

If the angle of impact has to be 45 degrees, the motion is parabolic and the cannon is above ground, the angle at interception of the parabola and cannon is less than 45 degrees.

This reasoning also tells you need steeper angle of launch if the cannon is below ground level.

Imagine a cannon at ground level that shoots a cannonball such that it passes through the location of the cliff-cannon at the required speed. This cannon is angled at 45''. When the cannonball passes through the location of the cliff-cannon, its angle of trajectory is <45''.

Assume the initial velocity is $u$ , shot angle be $\theta$ , initial height from the ground is $h>0$ , the time taken to hit the ground is $t$ , then we have $-h=ut\sin\theta-\frac12gt^2$ And then the range can be expressed as $R=ut\cos\theta$

Take derivative with respect to $\theta$ to the first equation, yields $\begin{aligned}0&=u(\sin\theta)\frac{dt}{d\theta}+ut\cos\theta-gt\frac{dt}{d\theta}\\\therefore\frac{dt}{d\theta}&=\frac{ut\cos\theta}{gt-u\sin\theta}\end{aligned}$

To find the extreme point on range, we should have: $\begin{aligned}\frac{dR}{d\theta}&=0\\u(\cos\theta)\frac{dt}{d\theta}-ut\sin\theta&=0\\\therefore\frac{dt}{d\theta}&=t\tan\theta\end{aligned}$

Link the two expressions of $d t/d\theta$ we have $\begin{aligned}\frac{ut\cos\theta}{gt-u\sin\theta}&=\frac{t\sin\theta}{\cos\theta}\\u\cos^2\theta&=gt\sin\theta-u\sin^2\theta\\u&=gt\sin\theta\\\sin\theta&=\frac u{gt}\end{aligned}$

But if we solve for $t$ in the expression of $h$ : $gt^2-2ut\sin\theta-2h=0$ , we will have one positive value and one negative value. Take positive as appropriate solution: $t=\frac{u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}}{g}>\frac{2u\sin\theta}{g}$

Then we have $\sin\theta=\dfrac{u}{gt}<\dfrac1{2\sin\theta}$ which leads to $\sin^2\theta<\dfrac12\Longleftrightarrow\sin\theta<\dfrac1{\sqrt2}$ .

This says that $\theta$ must less than $45^\circ$ .

Ps: I like to solve Physics problem mathematically...

I have a more mathematically worked out solution for anybody interested. Let $h_0$ be the initial height, $v_0$ be the initial velocity, $\theta$ is the angle between horizontal and cannon, $g$ be gravity and $t$ be time, then the following equations of motion hold:

$\begin{aligned} x &= v_0 \cos\theta t \\ y &= h_0 + v_0 \sin\theta t - \frac{1}{2}gt^2 \\ \end{aligned}$

Solving for $t$ in second equation and plugging into the first gives:

$0 = h_0 + x \tan \theta - \frac{gx^2}{2v_0^2} \sec^2 \theta$

Since $x = x(\theta)$ is a function of the angle, we can implicitly differentiate the above in terms of $\theta$ :

$0 = x'(\theta)\tan\theta + x\sec^2\theta - \frac{g}{2v_0^2} \bigg[ 2xx'(\theta)\sec^2\theta + 2x^2\sec^2\theta\tan\theta \bigg]$

We want the maximum angle so set $x'(\theta) = 0$ . Then the equation contains the max angle $\alpha$ and the max range $d$ :

$\begin{aligned} 0 &= d\sec^2\alpha - \frac{gd^2}{v_0^2}\sec^2\alpha\tan\alpha \\ 0 &= d\sec^2\alpha \bigg[ 1 - \frac{gd}{v_0^2}\tan\alpha \bigg] \\ \end{aligned}$

We don't want $d = 0$ and clearly $\sec^2\alpha \neq 0$ therefore:

$\begin{aligned} 0 &= 1 - \frac{gd}{v_0^2}\tan\alpha \\ d\tan\alpha &= \frac{v_0^2}{g} \\ \end{aligned}$

Now we can plug the above into the equation we derived above which relates $x$ and $\theta$ , rearranging a bit and manipulating it will yield:

$0 = h_0 + \frac{v_0^2}{g} - \frac{v_0^2}{2g} \csc^2\alpha$

Which we can rewrite as follows:

$\sin^2\alpha = \frac{v_0^2}{2(gh_0+v_0^2)}$

Using a trig identity we can finally write it as follows:

$\cos(2\alpha) = \frac{gh_0}{gh_0 + v_0^2}$

From there we can solve for $\alpha$ and so on... Here we can see that the maximum angle $\alpha$ depends on the initial height $h_0$ and the initial velocity $v_0$ so there is no definite answer for the maximum angle like the simple case. However we DO know that the angle is between 0 and 45 because

$\frac{gh_0}{gh_0 + v_0^2} > 0$

When this term IS zero, that is when the cannon is on the ground $h_0 = 0$ , we get the simple case:

$\cos(2\alpha) = 0$

Where we can clearly see that $\alpha = 45^o$

See the problem Maximizing Range 2 for mathematical solution.

When on the ground the canon is projected at an angle so that trajectory have both max height as well as distance and the angle in the case is 45° as it is in the projectile motion case. But at a height ,h if we say,the projected angle should be lower as the condition of achieving the max height is compensated by some factor by the height. And As speed is same for both canon balls,So in accordance with upper points the angle should be less than 45 in order to attain the same max distance and the same could be reversed if it was in depression like a hole.

Let H be the point we’re the cannon is located, and P is the point were the cannon reaches its maximum range. Then the cannon must be pointing 45 degrees from the line HP. If H is above ground then the angle theta must be lower than 45 degrees.

Gravity will act longer on the cannonball when shot from the cliff, so you should increase the cosine component of the force which is done by decreasing the angle.

Horizontally S = sinθ * V0 (initial speed) * t Assume the cannon is placed at a extremely high land, and 0<θ<90, t will be extremely large in this case. S will be max when θ is exetremely close to 0.

Logically, if there was no ground level, or the cliff was infinitely high, but you still wanted to launch as far as possible, it would be a waste to have any vertical initial velocity. So if the cliff was infinitely high, the optimal angle would be 0. As the height decreases, the angle increases, probably in some way relating to the $\tan$ function. Obviously this isn't rigorous, but it works for multiple choice.

As the canon is above the ground at a height, it has got some 'free altitude' to travel with the help of gravity and gain more horizontal distance than the one on the ground. The angle of launch splits the force imparted on the cannonball into horizontal and vertical components; at a 45-degree angle, the force splits equally and allows it to reach an optimal distance which is a trade-off between and horizontal reach and vertical climb.

Here, as the canon on the elevation gets an extra free altitude to travel, gravity takes care of it as it is coming down; So, to reach farther, more force has to be dedicated to the horizontal component; thus reducing the angle with respect to the 'free altitude' available.