1.6 Projectile Motion

Let’s now look at a concrete example of an experiment which will bring together everything we have looked at in this chapter. Suppose we have a cannon situated at the origin which shoots a projectile with a fixed initial velocity $\boldsymbol{v}_{0}$ which makes an initial angle $\theta$ relatitive to the ground. Can we work out how far the projectile will fly and what its flight time is?

1.6.1 Solving for the Motion

This is going to be a 2D problem as we have two axes of motion. We shall label the horizontal direction that the cannon is shooting along the $x$ axis and the vertical direction the $y$ axis. As stated in the problem, the cannon is located at the origin. We can now write the initial velocity as

\boldsymbol{v}_{0}=v_{0}\cos(\theta)\hat{\boldsymbol{\imath}}+v_{0}\sin(\theta% )\hat{\boldsymbol{\jmath}},

(1.45)

where $v_{0}=\lvert\boldsymbol{v}_{0}\rvert$ . Ignoring air resistance, there are no forces acting on the projectile in the $x$ direction. By Newton’s second law this means that $a_{x}=0$ and we can immediately write

x(t)=v_{0}\cos(\theta)t,

(1.46)

using the SUVAT equation 1.17. In the $y$ direction, the only force acting on the projectile is the constant force of gravity, so Newton’s second law tells us

F_{y}=-mg=ma_{y}.

(1.47)

Hence, we have

y(t)=v_{0}\sin(\theta)t-\frac{1}{2}gt^{2},

(1.48)

again by equation 1.17.

1.6.2 Flight Time and Range

Now, the time of flight $t_{f}$ will be given when $y(t_{f})=0$ . Solving for this, we get

$\displaystyle y(t_{f})=$	$\displaystyle v_{0}\sin(\theta)t_{f}-\frac{1}{2}gt_{f}^{2}=0$	(1.49)
	$\displaystyle v_{0}\sin(\theta)t_{f}=\frac{1}{2}gt_{f}^{2}$	(1.50)
	$\displaystyle t_{f}=\frac{2v_{0}\sin(\theta)}{g}.$	(1.51)

Let’s consider briefly if this answer makes physical sense. If the initial speed of the projectile $v_{0}$ was higher, then the flight time would be longer. Additionally, for a fixed initial speed, a projectile fired at a higher angle would have a longer flight time because more of the initial velocity was aimed along the vertical direction. On the other hand, if gravity was stronger, the projectile would fall to the ground faster and the flight time would be shorter.

Finally, the range of the projectile is given by

	$\displaystyle x_{f}=x(t_{f})$	$\displaystyle=v_{0}\cos(\theta)t_{f}$		(1.52)
		$\displaystyle=\frac{2v_{0}^{2}\sin(2\theta)}{g}.$		(1.53)

So the maximum range is given when $\theta=\frac{\pi}{4}$ .

Example 1.5.

A motorcyclist is doing a stunt jump between two buildings. If the buildings are separated by a distance $d$ and have a vertical height difference $h$ , what is the minimum velocity the motorcyclist needs to make the jump?