# What is the optimum angle of projection when throwing a stone off a cliff?

You are standing on a cliff at a height $h$ above the sea. You are capable of throwing a stone with velocity $v$ at any angle $a$ between horizontal and vertical. What is the value of $a$ when the horizontal distance travelled $d$ is at a maximum?

On level ground, when $h$ is zero, it's easy to show that $a$ needs to be midway between horizontal and vertical, and thus $\large\frac{\pi}{4}$ or $45°$. As $h$ increases, however, we can see by heuristic reasoning that $a$ decreases to zero, because you can put more of the velocity into the horizontal component as the height of the cliff begins to make up for the loss in the vertical component. For small negative values of $h$ (throwing up onto a platform), $a$ will actually be greater than $45°$.

Is there a fully-solved, closed-form expression for the value of $a$ when $h$ is not zero?

• I highly recommend throwing stones off a cliff when you're at the seaside. It's a lot of fun. This problem grew out of an argument I had with my family when we were at the seaside, that 45° isn't the best angle when you're on a cliff. Jul 23, 2010 at 14:16
• Should we take into account anatomic concerns? Because for example to throw a stone at a 45° angle, one have to release it quite early. Which would cause a lower speed compared to a later release on a lower angle. So we might have a speed/angle trade-off here. Also, the arm and wrist configuration should cause a non linear acceleration curve. (peaking at lower angles?) Aug 12, 2017 at 2:16

Assume no friction and uniform gravity g.

If you throw a stone at point (0, h), with velocity (v cos θ, v sin θ), then we get

\begin{align} d &= vt\cos\theta && (1) \\ 0 &= h + vt\sin\theta - \frac12 gt^2 && (2) \end{align}

The only unknown to be solved is t (total travel time). We could eliminate it by using $t = \frac d{v\cos\theta}$ to get

$$0 = h + d\tan\theta - \frac{gd^2\sec^2\theta}{2v^2}\qquad(3)$$

Then we compute the total derivative with respect to θ:

\begin{align} 0 &= \frac d{d\theta}\left(d\tan\theta\right) - \frac g{2v^2}\frac d{d\theta}\left(d^2\sec^2\theta\right) \\ &= \ldots \end{align}

and then set $\frac{dd}{d\theta}=0$ (because it is maximum) to solve d:

$$d = \frac{v^2}{g\tan\theta}$$

Substitute this back to (3) gives:

\begin{align} h &= \frac{v^2}g \left( \frac1{2\sin^2\theta} - 1\right) \\ \Rightarrow \sin\theta &= \left( 2 \left(\frac{gh}{v^2} + 1\right) \right)^{-1/2} \end{align}

This is the closed form of θ in terms of h.

• Wow, so you subbed the other way round. Very nice Jul 23, 2010 at 23:15
• Very elegant. Thank you. Sep 10, 2010 at 20:30
• ps. physics.stackexchange.com/a/23187/8446 gives an equivalent closed form for $θ$ in terms of $\tan$ instead of $\sin$. Apr 3, 2012 at 11:55

A projectile is thrown from a height $$h$$, with speed $$u$$ at an angle $$\theta$$ from the horizontal. With $$h,u$$ fixed we wish to find $$\theta$$ to maximise $$s$$, the horizontal distance covered before the projectile hits the ground.

Let $$\vec{u}, \vec{v}$$ be the initial and terminal velocities respectively. By conservation of energy we have: $$|\vec{u}| =u,\qquad\qquad |\vec{v}|=\sqrt{u^2+2gh},$$ so $$|\vec{u}\times \vec{v}|=u\sqrt{u^2+2gh}\left(\sin\alpha\right),\qquad (1)$$ where $$\alpha$$ is the angle between $$\vec{u}$$ and $$\vec{v}$$.

On the other hand $$\vec{u}=\left(\begin{array}{c}u\cos\theta\\0\\u\sin\theta\end{array}\right),\qquad \vec{v}=\left(\begin{array}{c}u\cos\theta\\0\\u\sin\theta-gt\end{array}\right),$$ where $$t$$ is the time that the projectile is in flight. Thus $$\vec{u}\times \vec{v}=\vec{u}\times (\vec{v}-\vec{u})=\left(\begin{array}{c}0\\gs\\0\end{array}\right).$$ Hence $$|\vec{u}\times \vec{v}|=gs. \qquad (2)$$

Combining $$(1)$$ and $$(2)$$ we get $$gs=u\sqrt{u^2+2gh}\left(\sin\alpha\right).\qquad (3)$$ Thus $$s$$ maximises at $$\frac ug \sqrt{u^2+2gh},$$ when $$\vec{u}$$ and $$\vec{v}$$ are perpendicular.

To find $$\theta$$ we can now use $$\vec{u}\cdot \vec{v}=0$$:$$u^2\cos^2\theta+u^2\sin^2\theta-gtu\sin\theta=0.$$ This simplifies to $$u^2=gs\tan\theta.$$ Using $$(3)$$ We conclude that $$\tan\theta=\frac u{\sqrt{u^2+2gh}}.$$

I don't have a complete solution, but I attempted to solve this problem using calculus.

$x'=v \cos a$
$y''= -g$ and (at $t=0) \quad y'= v \sin a$
So, $y'= v \sin a -gt$
$x_0=0$, so $x=vt \cos a$
$y_0=h$, so $y=vt \sin a - \frac12 gt^2+c$ (integrating with respect to $t$)
Subbing in $h, y=vt \sin a - \frac12 gt^2+h$

The ball will hit the ground when $y=0$.

This is as far as I got, but it appears that you can find a closed solution after all. I originally tried solving the quadratic for $t$ and subbing that into $x$, but it seems to work much better to do the substitution the other way round. I will leave this solution here in case anyone wants to see how to derive the basic equations for $x$ and $y$.

• I specifically excluded the case when h=0, because I know how to solve that. We know t isn't going to be zero, so we can set y=0, and get t = 2 v sin a / g, x = 2 v^2 sin a cos a / g = v^2 sin 2a / g. This is at a maximum when 2a = π and so a = π/2. Jul 23, 2010 at 14:05
• Also, you made a mistake and used cos a for y, although I know you meant sin a because you used this for y'. Jul 23, 2010 at 14:08
• @Neil: I meant to say subbing in y=0 to find when the ball hits the ground. You are right about the second comment too, I accidentally integrated with respect to t and then with respect to a as well. Jul 23, 2010 at 14:19
• I meant 2a = π/2 and so a = π/4 above. Jul 23, 2010 at 14:32