Question

A pirate ship 560 m from a fort defending a harbour entrance. A defence cannon is located at sea level, fires the ball at the initial speed \[82{m}/{s}\;\].
(i) At what angle \[\theta \]from horizontal must a ball be fired?
(ii) Maximum range of cannonball

Answer 1

Hint: The angle from the horizontal along which a ball must be fired can be calculated by substituting the given values in the formula used for calculating the range of the projectile. The maximum range can be found by substituting the angle for which the sin function gives maximum value.

Formula used:
\[R=\dfrac{{{u}^{2}}\sin (2\theta )}{g}\]

Complete answer:
From given, we have the data,
The horizontal range is, R = 560 m
The initial speed of the ball is, \[u=82{m}/{s}\;\].
The formula for calculating the range of the horizontal is as follows.
\[R=\dfrac{{{u}^{2}}\sin (2\theta )}{g}\]
Where u is the initial speed, g is the gravitational constant and \[\theta \]is the angle of inclination.
(i) The angle of inclination \[\theta \]from the horizontal
Rearrange the terms of the formula for obtaining the formula in terms of the angle of inclination.
Thus, we have,
\[\begin{align}
  & \sin (2\theta )=\dfrac{Rg}{{{u}^{2}}} \\
 & \Rightarrow 2\theta ={{\sin }^{-1}}\left( \dfrac{Rg}{{{u}^{2}}} \right) \\
 & \Rightarrow \theta =\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{Rg}{{{u}^{2}}} \right) \\
\end{align}\]
Now substitute the given values in the above equation to find the value of the angle of inclination.
\[\begin{align}
  & \theta =\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{Rg}{{{u}^{2}}} \right) \\
 & \Rightarrow \theta =\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{560\times 9.8}{{{82}^{2}}} \right) \\
 & \Rightarrow \theta =\dfrac{1}{2}{{\sin }^{-1}}(0.816) \\
\end{align}\]
Here we obtain the two values of the angle, as
\[\begin{align}
  & \Rightarrow {{\sin }^{-1}}(0.816)=54.7{}^\circ \\
 & \Rightarrow {{\sin }^{-1}}(0.816)=125.3{}^\circ \\
\end{align}\]

Now substitute these values of the angles in the above equation for the further calculation.
So, we get,
\[\begin{align}
  & \Rightarrow \theta =\dfrac{1}{2}(54.7{}^\circ )\approx 27{}^\circ \\
 & \Rightarrow \theta =\dfrac{1}{2}(125.3{}^\circ )\approx 63{}^\circ \\
\end{align}\]
At the angles of \[27{}^\circ \]or \[63{}^\circ \]from the horizontal, a ball must be fired.

(ii) Maximum range of cannon ball
The formula for calculating the range of the horizontal is given as follows.
\[R=\dfrac{{{u}^{2}}\sin (2\theta )}{g}\]
For a maximum range, the total angle should be \[90{}^\circ \], as the maximum value of sin function is obtained for this angle only.
So, we have,
\[\begin{align}
  & R=\dfrac{{{82}^{2}}\sin (2\times 45{}^\circ )}{9.8} \\
 & \Rightarrow R=\dfrac{6724\sin (90{}^\circ )}{9.8} \\
 & \Rightarrow R=\dfrac{6724}{9.8} \\
 & \Rightarrow R=686\approx 690\,m \\
\end{align}\]
The maximum range of the cannonball should be 690m.
The value of the angle \[\theta \]from horizontal a ball be fired is \[27{}^\circ \]or \[63{}^\circ \] and the maximum range of cannonball should be 690 m.

Note:
The things to be on your finger-tips for further information on solving these types of problems are: The angle in the formula should be multiplied by 2, so while calculating the maximum range, half the angle of the maximum angle should be considered. Even the units of the parameters should be taken into consideration.