# Physics with Calculus/Mechanics/Projectile Motion

## Projectile Motion

Using the equations we derived in the last section, we can now use them to model the motion of a projectile. A projectile is an object upon which the only force acting is gravity, which means that in all situations, the acceleration in the y direction, $a_y = -g$. For simplicity, we will assume that the path of a projectile, also called its trajectory, will always be in the shape of a parabola, and that the effect of air resistance upon the projectile is negligible.

The Horizontal Motion

Since the only force acting upon the object is gravity, in the y direction, there is no acceleration in the x direction.

Let us assume that the projectile leaves the origin at time t = 0 and with speed vi. Then we have a vector vi that makes an angle of θi with the x-axis. Then, using a bit of trigonometry, we have the following:

$\cos \theta_i = \frac{v_{xi}}{v_i}$

and

$\sin \theta_i = \frac{v_{yi}}{v_i}$

Rearranging for the initial velocities, we get the initial x and y components of velocity to be

$v_{xi} = v_i\cos \theta_i$ and $\ v_{yi} = v_i\sin \theta_i$
$x = v \cos(\theta) t$

The Vertical Motion

There is a constant acceleration down, g which is the force of gravity. Accelecration is the instantaneous rate of change of velocity so:

$\frac{dv}{dt} = a = g$

therefore we can integrate acceleration with respect to time to get velocity

$\int dv = \int g dt$
$v = gt$

Velocity is the instantaneous rate of change of displacement so:

$\frac{dd}{dt} = v$

We can also integrate velocity with respect to time to get displacement

$\int dx = \int v dt$
$\int dx = \int gt dt$
$d = \frac{1}{2} g t^2$