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 &theta;i 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
 * $$\int dx = \int v dt$$
 * $$\int dx = \int gt dt$$
 * $$d = \frac{1}{2} g t^2$$