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Chapter 2

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


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}

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

x=300 \sqrt{3} t

so that

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}.

This parabola is sketched in Figure 7 .