# Learn Projectile Motion (Again)

Projectile Motion is the most classic high school physics problem in existence. It might be a little bit boring, but it’s very important to master, and it’s a great way to apply what you’ve learned in kinematics. Here’s the set up.

First of all, what are the forces acting on the ball while it’s in the air? If you neglect air resistance (think baseball instead of balloon), then the only force is gravity! We know gravity always points downward, so this means that the vertical speed is constantly decreasing and the horizontal speed is constant.

Let’s write this out mathematically. Let $x(t)$ be the horizontal position of the ball and let $y(t)$ be the vertical position, both as functions of time $t$. Notice that if $y<0$, then the ball is lower than where it started (which is possible if you’re throwing off a cliff for example). Let $v_{x}(t)$ be the horizontal speed and $v_{y}(t)$ be the vertical speed.

## Horizontal component of projectile motion

Since $v_{x}(t)$ doesn’t change, $v_{x}(t) = v_{x}(0)$. By trigonometry, the initial horizontal speed is just $v_{x}(0) = v_{0} \cos{\theta}$. See the image below.

Therefore for all times $t$, $v_{x}(t) = v_{0} \cos{\theta}$.

Since this is a constant speed, the position $x(t) = v_{x} t = v_{0} \cos{\theta} t$.

This is a pretty cool result. Let’s say that

$v = 1 \, m/s$

and

$\theta = 60^{\circ}$.

After $1 \, s$, how far horizontally has the ball travelled? You can just plug into the above equation without doing anymore work at all. It’s just

$x(1 \, s) = 1 \, m/s \cdot \cos{60^{\circ}} \cdot 1 \, s$.

Recall that

$\cos{60^{\circ}} = 1/2$,

so

$x(1 \, s) = 0.5 \, m$.

### Vertical component of projectile motion

That was fun, but how about $y$? Since $v_{y}$ is decreasing at a constant rate of $g = 9.8 \, m/s^{2}$ downwards, after a time $t$, $v_{y}$ decreases by $gt$. So

$\displaystyle v_{y}(t) = v_{y}(0)-gt$

What’s $v_{y}(0)$? Using trigonometry, it’s just $v_{0} \sin{\theta}$! So

$\displaystyle v_{y}(t) = v_{0} \sin{\theta} - gt$

### IMPORTANT TIMES

Like before, you can do a lot with this equation alone. For example, how long does it take for the ball to reach it’s maximum height? The key insight is that at the maximum point, the vertical speed of the ball is zero! If we let this time be $T_{1/2}$ (the funny notation will be explained in a moment), then plugging into the equation we just derived,

$\displaystyle v_{y}(T_{1/2}) = 0 = v_{0} \sin{\theta} - gt_{1/2} \Longrightarrow T_{1/2} = \frac{v_{0}\sin{\theta}}{g}$.

How long does it take for the ball to hit the ground? We can use conservation of energy. If we measure gravitational energy relative to the ground, then the initial energy is all kinetic: Just before it hits the ground, the gravitational energy is zero again, so the energy is also purely kinetic. This means that the speed is the same. Since the horizontal component of speed is constant, the vertical speed must have the same magnitude as the initial vertical velocity. But since it’s now moving down, the $v_{y}$ switches signs. Therefore if the total air time is $T$, then

$\displaystyle v_{y}(T) = -v_{0} \sin{\theta} \Longrightarrow -v_{0} \sin{\theta} = v_{0}\sin{\theta} - gT \Longrightarrow T = \frac{2v_{0} \sin{\theta}}{g}$

Notice this is twice the time it takes to reach it’s maximum height, hence the notation $T_{1/2}$.

Using this equation, we can find the total distance travelled by the ball. It’s actually really easy! The horizontal velocity is constant, so the total distance is

$\displaystyle d_{tot} = v_{0} \cos{\theta} T = \frac{2v_{0} \sin{\theta} \cos{\theta} }{g}$

If you’ve done some trigonometry, then you’ll recognize

$\displaystyle 2\sin{\theta}\cos{\theta} = \sin{2\theta}$. Therefore

$\displaystyle d_{tot} = \frac{v_{0} \sin{2\theta}}{g}$

What value of $\theta$ maximizes the total distance? Well $\sin{2 \theta} \leq 1$, so the maximum distance is

$\displaystyle d_{max} = \frac{v_{0}}{g}$

This occurs exactly when $2 \theta = 90^{\circ} \Longrightarrow \theta = 45^{\circ}$. However, if you’re actually throwing off a cliff (instead of on flat ground), then the optimal value of $\theta$ actually decreases.

How about the position of the ball? This is a bit harder, but there’s a formula for it. According to this formula, if you’re moving with an initial speed $u$ and you’re accelerating forward at a rate $a$, then your position relative to your initial position is

$\displaystyle d = ut+\frac{1}{2}at^2$

If you’re not familiar with this formula, I’ll make a post on it soon. Since the initial vertical speed is up and we’re accelerating down, the $a$ we use is actually negative!

$\displaystyle a = -g$

Therefore

$\displaystyle y(t) = v_{0} \sin{\theta}t - \frac{1}{2}gt^{2}$

So for any time $t$, you can find the vertical position. What’s the shape of the curve? Since

$\displaystyle x = v_{0} \cos{\theta} t \Longrightarrow t = \frac{x}{v \cos{\theta}}$, we can substitute this into the above equation to get

$\displaystyle y = v_{0} \sin{\theta} \frac{x}{v \cos{\theta}} - \frac{1}{2}g(\frac{x}{v \cos{\theta}})^{2}$

If you’ve studied quadratic equations before, you’ll recognize this as the equation of an upside down parabola. Here’s what it looks like.

There’s a nice little trick that might give you some good intuition about projectile motion. Imagine that you’re launching the projectile and Patricia is moving to the right with a speed $v_{0} \cos{\theta}$. Then Patricia is moving at the same horizontal speed as the ball, so if you ask her, she’ll tell you that the ball is moving straight up and down. The initial speed of the ball is just the vertical component of the initial velocity in your frame. This gives you some intuition about why total air time only depends on vertical speed.

There’s another interesting thing you should know about projectiles (although this is actually true about any path taken by a particle under a time independent potential). If you throw a ball and while it’s in the air, you suddenly reverse the direction of the velocity, then the ball will follow the exact same path except with a reversed velocity at each point. In other words, you get the same thing as if you record a video of the projectile and then rewind the video. This is known as time reversal symmetry and it is a property of Newton’s laws. In particular, if you throw a ball to your friend, then for your friend to throw to you, he can just reverse the velocity of the ball when he catches it.