Question 1.6: A cannonball shot from a cannon has an initial velocity of 6...

We place the x – and y-axes so that the mouth of the cannon is at the origin (see Figure 6). The velocity vector $\mathbf{v}$ can be resolved into its vertical and horizontal components:

A unit vector in the direction of $\mathbf{v}$ is

So initially,

The scalar $|\mathbf{v}|$ is called the speed of the cannonball. Thus initially, $v_{x}=300 \sqrt{3} \mathrm{~m} / \mathrm{sec}$ (the initial speed in the horizontal direction) and $v_{y}=300 \mathrm{~m} / \mathrm{sec}$ (the initial speed in the vertical direction). Now the vertical acceleration (due to gravity) is

$a_{y}=-9.81 \mathrm{~m} / \mathrm{sec}^{2} \quad \text { and } \quad v_{y}=\int a_{y} d t=-9.81 t+C$.

Since, initially, $v_{y}(0)=300$, we have $C=300$ and

Then

$y(t)=\int v_{y} d t=-\frac{9.81 t^{2}}{2}+300 t+C_{1}$.

and since $y(0)=0$ (we start at the origin), we find that

To calculate the x-component of the position vector, we note that, ignoring air resistance, the velocity in the horizontal direction is constant;${ }^{\dagger}$ that is,

Then $x(t)=\int v_{x} d t=300 \sqrt{3} t+C_{2}$, and since $x(0)=0$, we obtain

Thus the position vector describing the location of the cannonball is

To obtain the Cartesian equation of this curve, we start with

so that

and

$y=300\left(\frac{x}{300 \sqrt{3}}\right)-\frac{9.81}{2} \frac{x^{2}}{(300 \sqrt{3})^{2}}=\frac{x}{\sqrt{3}}-\frac{9.81}{540,000} x^{2}$.

This parabola is sketched in Figure 7 .