Projectile motion describes the path of an object launched in the vertical plane with an initial velocity, after which it is subject only to gravity.
Air resistance is ignored.
Horizontal and vertical motion are independent.
Horizontal acceleration = 0.
Vertical acceleration = \(g\) downward.
A cannon ball is fired from a cannon with an initial velocity Vo m/s2
\[
\vec{v}_{o}
= \vec{v}_{o}\,\mathbf{i}
+ \vec{v}_{o}\,\mathbf{j}
\]
\[
v_{ox} = v_{o}\cos\theta
\]
\[
v_{oy} = v_{o}\sin\theta
\]
The position and velocity vectors change during flight, but the projectile always experiences:
- constant downward acceleration \(g\)
- zero horizontal acceleration
The horizontal displacement (x - x0) can be found using
\[ s = ut + \frac{1}{2}at^{2} \]
so
\[
x - x_{o}
= v_{ox} t + \frac{1}{2} a t^{2}
\]
Since there is no horizontal acceleration:
\[
x - x_{o} = v_{ox}\, t
\]
and
\[
x - x_{o}
= \left( v_{o}\cos\theta \right) t
\]
Vertical displacement (y - y0):
\[
y - y_{o}
= v_{oy} t + \frac{1}{2} a t^{2}
\]
\[
y - y_{o}
= v_{oy} t - \frac{1}{2} g t^{2}
\]
so
\[
y - y_{o}
= \left( v_{o}\sin\theta \right)t
- \frac{1}{2} g t^{2}
\]
Equation of the Trajectory
\[
x - x_{o}
= \left( v_{o}\cos\theta \right)t
\]
\[
t = \frac{x - x_{o}}{\,v_{o}\cos\theta\,}
\]
substituting for t:
\[
y - y_{o}
= v_{o}\sin\theta
\left(
\frac{x - x_{o}}{\,v_{o}\cos\theta\,}
\right)
- \frac{1}{2} g
\left(
\frac{x - x_{o}}{\,v_{o}\cos\theta\,}
\right)^{2}
\]
Using initial conditions,
\(
x_{o} = 0
, y_{o} = 0
\)
\[
y
= v_{o}\sin\theta
\left(
\frac{x}{\,v_{o}\cos\theta\,}
\right)
- \frac{1}{2} g
\left(
\frac{x}{\,v_{o}\cos\theta\,}
\right)^{2}
\]
so
\[
y
= \left(
\frac{x\,v_{o}\sin\theta}{\,v_{o}\cos\theta\,}
\right)
- \frac{g x^{2}}{\,2\left( v_{o}\cos\theta \right)^{2}}
\]
\[
= x\tan\theta
- \frac{g x^{2}}{\,2\left( v_{o}\cos\theta \right)^{2}}
\]
This is the equation of the projectile’s path.
The range \(R\) is the horizontal distance travelled when the projectile returns to its launch height.
Since the original launch height is yo=0,
\[
y
= x\tan\theta
- \frac{g x^{2}}{\,2\left( v_{o}\cos\theta \right)^{2}}
\]
\[
0
= x\tan\theta
- \frac{g x^{2}}{\,2\left( v_{o}\cos\theta \right)^{2}}
\]
\[
x \tan\theta
= \frac{g x^{2}}{\,2\left( v_{o}\cos\theta \right)^{2}}
\]
\[
2\left( v_{o}\cos\theta \right)^{2}\tan\theta
= \frac{g x^{2}}{x}
\]
\[
x
= \frac{2\left( v_{o}\cos\theta \right)^{2}\tan\theta}{g}
\]
\[
= \frac{2\,v_{o}^{2}\cos^{2}\theta}{g}
\left(
\frac{\sin\theta}{\cos\theta}
\right)
\]
\[
= \frac{2\,v_{o}^{2}\cos\theta\,\sin\theta}{g}
\]
but \(
\sin 2\theta = 2\sin\theta\cos\theta
\)
\[
x
= \frac{v_{o}^{2}\,\sin 2\theta}{g}
\]
also
\[
x - x_{o} = R
\]
so
\[
R
= \frac{v_{o}^{2}\,\sin 2\theta}{g}
\]
Maximum range occurs when \(\sin2\theta = 1\),
so the projectile travels furthest when launched at 45°.
