Parabolic Motion

We all know from algebra what a parabola is. It's the graph of a quadratic function (a polynomial of degree $2$). However, it is more than just a mathematical graph in the world of physics. It turns out that when an object is given an initial velocity at any particular direction, if it's not on the ground, it follows the trajectory of a parabola!

Let us try to discover the path of the trajectory ourselves. Let us say we are on flat ground and a soccer ball is kicked up at angle $\theta$ with respect to the ground, with velocity $v$. Let the point of impact be the origin on the coordinate plane. Then, since there is no horizontal acceleration, the $x$-position of the ball is given by $$x = v_x t \implies x = vt \cos \theta.$$

Now, remember your kinematics! If we let upwards be the positive direction, we find that if the acceleration of gravity is $g$, we get $$y = v_y t - \frac {g}{2} t^2 \implies y = vt \sin \theta - \frac {g}{2} t^2.$$

From here, it is easy to substitute and solve to see that the $xy$-equation is a parabola, unless air resistance is present to spoil it! In future sections, you can read more about the properties of these parabolic trajectories, which are given the name projectile motion.

Parabolic Motion

Universal Truths for Projectile Motion:

1. $V_x$ = constant $V_x$ = Velocity in the x-direction

1. $V_y$ = $V_{oy} - at$ $V_y$ = Velocity in the y-direction, $V_{oy}$ = Initial Velocity in the y-direction

1. $a_x$ = 0$m/s^2$ $a_x$ = Acceleration in the x-direction

1. $a_y$ = $a_g$ = -9.8$m/s^2$ $a_g$ = Acceleration due to gravity

1. $X$ = $V_xt_{air}$ $X$ = Displacement in x-direction

1. $X_y$ = $V_{oy}t_{air} + \frac{1}{2}at^2$ $X_y$ = Displacement in y-direction, $t_{air}$ = Time object is in the air

1. $R$ = $V_xt_{air}$ $R$ = Range ($m$)

Universal Truths for PARABOLIC Motion:

1. $R_{max} = 45^\circ$ You achieve the maximum range if you shoot a projectile is shot at a $45^\circ$ angle (ignoring air resistance)

Link to image above

Test yourself!:

Use kinematic equations to solve this problem. Use -10$m/s^2$ instead of -9.8$m/s^2$ for $a_g$

You shoot a projectile at a $60^\circ$ angle with an $V_o$ of 141.4213 $m/s$.

$t_{max}$ = Time when the projectile is at the highest point in the arc

$h_{max}$ = Height of the projectile at the highest point in the arc

Is projectile motion still parabolic motion with drag?

It is well-known that the trajectory of the projectile motion is symmetrical. But what if we include drag, will the trajectory still be symmetrical?

An obvious answer here is no, simply because drag destroys the symmetry in the equations of motion. The forces acting on the projectile when it is moving upwards and when it is moving downwards are no longer the same so it will not be parabolic motion.

Thinking Question: If it not symmetrical, which side is the peak of the trajectory closer to?