# The angle of launching an Angry Bird that will land the bird furthest from the launching point is 45 degrees?

This is a question motivated by Angry Birds.

When playing the game, I notice that if the initial velocity is constant, the way to land a bird furthest away from the launching point is by launching the bird at 45 degrees from the ground.

Is it possible to obtain a derivation and a proof of this?

45 degrees is, in fact, the angle for maximum range for a projectile with no air resistance. In the absence of air resistance, the only force acting is gravity, which causes a constant acceleration of g downwards. this determines the amount of time the particle spends in the air, via the formula for the position of a particle with constant acceleration:

$y(t) = y(0) + v_y t + \frac{1}{2}a_y t^2$

Putting in the relevant parameters (start and end positions both 0, acceleration -g (negative because it's downward)) this becomes:

$0 = v_y t - \frac{1}{2}g t^2$

which we solve to get:

$t = \frac{2v_y}{g}$

This time then goes into the equation for the horizontal position:

$x(t) = x(0) + v_x t + \frac{1}{2}a_x t^2$

As there's no horizontal force acting, this reduces to just

$x(t) = v_x t = \frac{2v_x v_y}{g}$

To get this in terms of the angle, we use the fact from trigonometry that for a velocity $v$ at an angle $\theta$ from the horizontal, the vertical velocity is $v \sin \theta$ and the horizontal velocity is $v \cos \theta$, giving us:

$x = \frac{2v \cos \theta v\sin \theta}{g} =\frac{v^2}{g}\sin 2\theta$

This has its maximum value for $\theta = \frac{\pi}{4}$, namely, 45 degrees from the horizontal.

• You should use $a_x$ and $a_y$ to make your answer clearer Commented Dec 29, 2010 at 15:21
• I generally try to avoid the proliferation of needless subscripts-- I find people get hung up on the meaning of those more often than they confuse the acceleration between dimensions. But that's probably less of a problem with this crowd, so I added them. Commented Dec 29, 2010 at 15:27
• @Chad Orzel you should note this is the case where the starting and ending positions are the same. I believe if I remember correctly that for a higher starting position (as in the game) an angle of 42 degrees produces the furthest horizontal travel. Commented Dec 29, 2010 at 17:53
• @msarchet: that would depend on the height difference. (I've heard that 42 degrees gives the furthest range for no height difference under realistic air resistance conditions) Commented Dec 29, 2010 at 20:33
• @David yea that's what I was thinking about when I wrote that. I had that angle tucked in the back of my head and couldn't remember where it came from. And I actually redid the math and it's always 45 degrees independent of the height since one time is negative (that's assuming ending height is always 0). Commented Dec 30, 2010 at 15:03

Marek asks for a "simpler conceptual argument"; there is one, using the basic observation that the distance travelled is linear in both the horizontal and vertical components of the initial velocity. It's not hard to fill in the rest, but I'm on a cross-country flight right now so it will need to wait a few hours.

Edit Ok, details:

The distance traveled is given by $d = v_xt$. One can either show with some algebra (as in Chad Orzel's answer) or otherwise that the flight time is linear in the vertical component: $t = cv_y$. Thus, $d$ is proportional to $v_xv_y$.

So we maximize the distance travelled by maximizing $v_xv_y$ subject to the constraint that the magnitude of the velocity vector is fixed: $|v| = c_2$. There's a bunch of ways to see how to do this; you can use calculus, but I prefer to observe that the level sets of $f(x,y) = xy$ are hyperbolas with asymptotes on the $x$ and $y$ axes, so the maximum must occur when $v_x = v_y$.

Even cuter is to appeal to the fact that both the constraint and the equation to maximize are quadratic, so there is either a unique maximum or the maximum occurs at one or both of the endpoints. Since the distance travelled at the endpoints is obviously zero, there is a unique maximum. By symmetry, it must occur when $v_x = v_y$.

• Once you get to $v_xv_y$ and $|v|=c_2$ you can find the solution by geometrical means. $v_xv_y$ is proportional to the area of a right triangle with $v_x$, $v_y$, $v$ as sides and $\theta$, the shooting angle as one of the angles. In turn, this area is proportional to $c_2 \times \sin(2\theta)$. Therefore it is maximised by a 45 degree angle. Commented Jan 4, 2011 at 2:29
• @Sklivvz: another very cute way to do it. Commented Jan 4, 2011 at 2:41

I just want to rederive the formula in easier way. That's because I want to minimize pencil pushing and love to talk a lot. We understand the problem much more easily that way.

$distance= v_x T$

You see that $v_x$ is constant.

So basically $v_x = v_{0x}$ where $v_{0x}$ is just the starting velocity on x axis. That's why the formula for the distance is so simple. $T$ which is the time taken can also be computed as easily.

where $v_x$ is the horizontal velocity and $T$ is total time taken.

$v_x = v\: cos\left ( \theta \right )$

and $T$ can be easily computed by knowing that the total time that it takes from launch to land.

But that's easy to compute. The kinetic energy at launch is the same with the kinetic energy at landing. Why? The potential energy is the same. Same height. The total energy is constant (if we ignore friction). So the kinetic energy is the same. So $|v|$ is the same. And $v_x$ is the same too. So $|v_y|$ is the same too. Only the direction is in reverse. At launch, $v_y = v_{0y}$. When landing $v_y = -v_{0y}$. So, during the process of launching the cannons or bullets or angry birds, $v_y$ have changed by $2v_{0y}$.

Here, $v_{0y}$ is simply the initial vertical upward velocity. Unlike $v_x$ that's trivially constant, $v_y$ changes a lot. But it doesn't matter because we only need to know the total time of "flying"

$T$ is just the time for $g$ to reverse $v_y$. Our gravitational acceleration will require $.5T$ to deplete all $v_0y$ and another $.5T$ to make $v_y$ becoming $-v_{0y}$

So, what is $T$? Well, it's simply the velocity difference, which is $2v_{0y}$ divided by $g$.

So $T=\frac{2v_{0y}}{g}$

Easy right?

Hence, $distance=v_x T = 2\frac{v_xv_y}{g}$

Obviously simple trigonometry will show that it is

$distance = 2\frac{v\: cos\left ( \theta \right )v\: sin\left ( \theta \right )}{g} = \frac{2v^2\, sin(\theta )cos(\theta )}{g}$

Someone will recognize that $2\, sin(\theta )cos(\theta )$ is simply $sin(2\theta)$. So we got the formula $distance = \frac{v^2\, sin(2\theta )}{g}$

that is maximized when $2\theta$ is 90 degrees or half a radian. That means when $\theta=45^0$

• no differential equation. not much pencil pushing. is this more understandable? that's how I teach my self physics. it's the only way people should learn physics and math. if only i can do this bigger Commented Sep 28, 2016 at 11:15