Question

An enemy ship is at a horizontal distance \[180\sqrt 3 \,{\text{m}}\] from a security cannon having a muzzle velocity \[60\,{\text{m/s}}\](\[g = 10\,{\text{m/}}{{\text{s}}^2}\])
A. Angle of elevation of cannon to hit ship is 30 and 60
B. Time of flight can be \[6\,{\text{s}}\]
C. Time of flight can be \[10\,{\text{s}}\]
D. Distance that the ship should be moved away from its initial position so that it becomes beyond the range of the cannon is \[48.6\,{\text{m}}\]

Answer 1

Hint:Use the formulae for horizontal range of the projectile and time of flight of the projectile. Using these two formulae, calculate the angle of projection of the cannon to hit the ship and time of flight of the cannon. Calculate the maximum range of the ship and subtract the given range of the ship from the maximum range to calculate the distance by which the ship should be moved.

Formulae used:
The horizontal range \[R\] of a projectile is
\[R = \dfrac{{{u^2}\sin 2\theta }}{g}\] …… (1)
Here, \[u\] is initial velocity of the projectile, \[\theta \] is angle of projection and \[g\] is acceleration due to gravity.
The time of flight \[T\] of a projectile is
\[T = \dfrac{{2u\sin \theta }}{g}\] …… (2)
Here, \[u\] is initial velocity of the projectile, \[\theta \] is angle of projection and \[g\] is acceleration due to gravity.

Complete step by step answer:
We have given that the horizontal range for the cannon upto the enemy ship is \[180\sqrt 3 \,{\text{m}}\].
\[R = 180\sqrt 3 \,{\text{m}}\]
The velocity of projection of the cannon is \[60\,{\text{m/s}}\].
\[u = 60\,{\text{m/s}}\]
Let us first calculate the angle of projection for the cannon. Substitute \[180\sqrt 3 \,{\text{m}}\] for \[R\], \[60\,{\text{m/s}}\] for \[u\] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in equation (1).
\[180\sqrt 3 \,{\text{m}} = \dfrac{{{{\left( {60\,{\text{m/s}}} \right)}^2}\sin 2\theta }}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow \sin 2\theta = \dfrac{{\left( {180\sqrt 3 \,{\text{m}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)}}{{{{\left( {60\,{\text{m/s}}} \right)}^2}}}\]
\[ \Rightarrow \sin 2\theta = \dfrac{{\sqrt 3 }}{2}\]
\[ \Rightarrow 2\theta = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow 2\theta = 60^\circ ,120^\circ \]
\[ \Rightarrow \theta = 30^\circ ,60^\circ \]
Hence, the angle of elevation of the cannon to hit the ship is \[30^\circ \] and \[60^\circ \].Hence, the option A is correct.

Let us now calculate time of flight for the angle of projection \[30^\circ \].Substitute \[60\,{\text{m/s}}\] for \[u\], \[30^\circ \] for \[\theta \] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in equation (1).
\[T = \dfrac{{2\left( {60\,{\text{m/s}}} \right)\sin 30^\circ }}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow T = \dfrac{{2\left( {60\,{\text{m/s}}} \right)\left( {\dfrac{1}{2}} \right)}}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow T = 6\,{\text{s}}\]
Thus, the time of flight of the cannon ball is \[6\,{\text{s}}\].Hence, the option B is correct.

Let us now calculate time of flight for the angle of projection \[60^\circ \]. Substitute \[60\,{\text{m/s}}\] for \[u\], \[60^\circ \] for \[\theta \] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in equation (1).
\[T = \dfrac{{2\left( {60\,{\text{m/s}}} \right)\sin 60^\circ }}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow T = \dfrac{{2\left( {60\,{\text{m/s}}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow T = 10.4\,{\text{s}}\]
\[ \Rightarrow T \approx 10\,{\text{s}}\]
Thus, the time of flight of the cannon ball is \[10\,{\text{s}}\].Hence, the option C is correct.

The range of the ship will be maximum when the angle of projection is \[45^\circ \].
Substitute \[45^\circ \] for \[\theta \], \[60\,{\text{m/s}}\] for \[u\] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in equation (1).
\[{R_{\max }} = \dfrac{{{{\left( {60\,{\text{m/s}}} \right)}^2}\sin 2\left( {45^\circ } \right)}}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow {R_{\max }} = \dfrac{{{{\left( {60\,{\text{m/s}}} \right)}^2}\sin 90^\circ }}{{10\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow {R_{\max }} = 360\,{\text{m}}\]
Hence, the maximum range of the cannon is \[360\,{\text{m}}\].

For the ship to be out of reach of the cannon, the distance between the enemy ship and cannon should be increased by a value \[x\] equal to maximum range of the cannon and present range of the cannon.
\[x = {R_{\max }} - R\]
Substitute \[360\,{\text{m}}\] for \[{R_{\max }}\] and for \[R\] in the above equation.
\[x = \left( {360\,{\text{m}}} \right) - \left( {180\sqrt 3 \,{\text{m}}} \right)\]
\[ \Rightarrow x = 180\left( {2 - \sqrt 3 } \right)\]
\[ \Rightarrow x = 48.6\,{\text{m}}\]
Thus, the distance by which the ship should be moved away from its initial position is \[48.6\,{\text{m}}\].Hence, the option D is correct.

Hence, the correct options are A, B, C and D.

Note:The students should be careful while calculating the angle of elevation of the cannon in order to hit the ship. The students should check all the possible values of the angle for the resulting value of sine of the angle. If we use only one value of the angle then the result will be incorrect. Hence, we should check all the values of the angle.