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, ${\displaystyle a_{y}=-g}$. 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 v_i. Then we have a vector v_i that makes an angle of θ_i with the x-axis. Then, using a bit of trigonometry, we have the following:

${\displaystyle \cos \theta _{i}={\frac {v_{xi}}{v_{i}}}}$

and

${\displaystyle \sin \theta _{i}={\frac {v_{yi}}{v_{i}}}}$

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

${\displaystyle v_{xi}=v_{i}\cos \theta _{i}}$ and ${\displaystyle \ v_{yi}=v_{i}\sin \theta _{i}}$

${\displaystyle x=v\cos(\theta )t}$

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:

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

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

${\displaystyle \int dv=\int gdt}$

${\displaystyle v=gt}$

Velocity is the instantaneous rate of change of displacement so:

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

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

${\displaystyle \int dx=\int vdt}$

${\displaystyle \int dx=\int gtdt}$

${\displaystyle d={\frac {1}{2}}gt^{2}}$