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 .