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## Q. 1.6

A cannonball shot from a cannon has an initial velocity of 600 m/sec. The muzzle of the cannon is inclined at an angle of 30°. Ignoring air resistance, determine the path of the cannonball.

## Verified Solution

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:

$\mathbf{v}=v_{x} \mathbf{i}+v_{y} \mathbf{j}$

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

$\mathbf{u}=\left(\cos 30^{\circ}\right) \mathbf{i}+\left(\sin 30^{\circ}\right) \mathbf{j}=\frac{\sqrt{3}}{2} \mathbf{i}+\frac{1}{2} \mathbf{j}$

So initially,

$\mathbf{v}=|\mathbf{v}| \mathbf{u}=600 \mathbf{u}=300 \sqrt{3} \mathbf{i}+300 \mathbf{j}$

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

$v_{y}=-9.81 t+300$

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

$y(t)=-\frac{9.81 t^{2}}{2}+300 t$

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,

$v_{x}=300 \sqrt{3} \mathrm{~m} / \mathrm{sec}$

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

$x=300 \sqrt{3} t$

Thus the position vector describing the location of the cannonball is

$\mathbf{s}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}=300 \sqrt{3} t \mathbf{i}+\left(300 t-\frac{9.81 t^{2}}{2}\right) \mathbf{j}$

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