Simple projectile motion

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Projectile motion happens in two directions: the x and the y axis. To understand it, we need to understand how the velocity of the projectile changes over time. On the x axis, $v_x$ doesn’t change because there’s no force and therefore no acceleration applied to it. However, on the y axis, $v_y$ changes because gravity drags the object down. Let’s see what the velocity is ⬇️

$$ \begin{cases} v_x=v_{0x}=v_0\cos(\theta)

v_y=v_{0y}-at=v_0\sin(\theta)-gt \end{cases} $$

From these formulas, we can see that the motion on the x axis is a uniform motion, and the one on the y axis is a uniformly accelerated motion. Their laws are writtyen like this ⬇️

$$ \begin{cases} x=v_xt+x_0

y=-\frac{1}{2}gt^2+v_{0y}t+y_0 \end{cases} $$

Where $g=9.81m/s^2$ (acceleration due to the force of gravity), $x_0$ is the x-coordinate of the starting point and $y_0$ is the y-coordinate of the starting point. From these two equations, we can find the equation of the trajectory that the ball draws in the air ⬇️

$$ t=\frac{x-x_0}{v_x}

y=-\frac{1}{2}g(\frac{x-x_0}{v_x})^2+\frac{v\sin\theta}{v\cos\theta}(x-x_0)+y_0=

=-\frac{g}{2v_x^2}x^2+(\tan\theta-\frac{2x_0}{v_x^2})x+\frac{x_0^2}{v_x^2}-x_0\tan\theta+y_0 $$

The equation above can be simplified if $x_0=0$:

$$ y=-\frac{g}{2v_x^2}x^2+x\tan\theta+y_0 $$

There are three main things tests might ask you about projectile motion:

time of flight, which is the time the projectile spends in the air. You can find it by finding the time at which the projectile touches the ground (i.e: $y=0$);
range, which is the distance covered by the projectile on the x axis during the time of flight. Use the time of flight $t$ to find $\Delta x=v_xt$;
maximum height, which is the highest the projectile goes before turning and going back to hit the ground. It corresponds to the moment at which the y axis velocity changes its sign and its direction (i.e: $v_y=0$). You can put the time that satisfies this requirement inside the y equation to find the highest point of the motion.

Projectile motion on an incline

A more complex situation to understand is that of a projectile motion on an incline. However, you will see that almost nothing changes compared to before, because it works just the same! There are two ways to approach a situation like this: either we change our frame of reference to be aligned with the incline, or we keep the standard one reminding that the projectile won’t hit the ground but the incline. Both systems work perfectly fine. I usually prefer to change the frame of reference when the maximum distance is requested in the problem, because it’s very easy to find if the frame of reference has the y axis perpendicular to the incline. As with simple projectile motion, in that case $v_y=0$, because the projectile then turns and goes on to hit the incline. However, if we kept the ground as our reference, we wouldn’t be able to use this because, as you can see from the image above, the projectile goes up relatively to the ground even after its maximum distance from the incline.

Let’s see how the equations for our motion change if the frame of reference changes. Acceleration is split into its two components:

$$ \begin{cases} a_x=g\sin(\theta)

a_y=g\cos(\theta) \end{cases} $$