The cannon on a battleship can fire a shell a maximum distance of\(32.0{\rm{ km}}\).

(a) Calculate the initial velocity of the shell.

(b) What maximum height does it reach? (At its highest, the shell is above\(60\% \)of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.)

(c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is \(6.37 \times {10^3}{\rm{ km}}\). How many meters lower will its surface be \(32.0{\rm{ km}}\) from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

When a projectile is launched with a certain initial velocity at a certain angle to the horizontal, the gravitational force of the Earth exerts a constant downwards acceleration on it, causing the projectile to travel in a curved parabolic path called a trajectory, and this motion is known as projectile motion.

The horizontal range of the projectile motion is,

\(R = \frac{{{u^2}\sin \left( {2\theta } \right)}}{g}\)

Here\(R\) is the horizontal range,\(u\) is the initial velocity,\(\theta \) is the angle of projection (for maximum range ) and\(g\) is the acceleration due to gravity.

Rearranging the above equation to get an expression for the initial velocity.

\(u = \sqrt {\frac{{Rg}}{{\sin \left( {2\theta } \right)}}} \)

Substitute\(32.0{\rm{ km}}\) for\(R\),\(9.8{\rm{ m}}/{{\rm{s}}^2}\) for\(g\), and\(45^\circ \) for\(\theta \),

\(\begin{aligned}{c}u = \sqrt {\frac{{\left( {32.0{\rm{ km}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}{{\sin \left( {2 \times 45^\circ } \right)}}} \\ = \sqrt {\frac{{\left( {32.0{\rm{ km}}} \right) \times \left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}{{\sin \left( {2 \times 45^\circ } \right)}}} \\ = 560{\rm{ m}}/{\rm{s}}\end{aligned}\)

Hence, the initial velocity of the shell is \(560{\rm{ m}}/{\rm{s}}\).

The maximum height attained by the projectile is,

\(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

Substitute\(560{\rm{ m}}/{\rm{s}}\) for\(u\),\(9.8{\rm{ m}}/{{\rm{s}}^2}\) for\(g\), and\(45^\circ \) for\(\theta \),

\(\begin{aligned}{}H = \frac{{{{\left( {560{\rm{ m}}/{\rm{s}}} \right)}^2} \times {{\sin }^2}\left( {45^\circ } \right)}}{{2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = \left( {8000{\rm{ m}}} \right) \times \left( {\frac{{1{\rm{ km}}}}{{{{10}^3}{\rm{ m}}}}} \right)\\ = 8.0{\rm{ km}}\end{aligned}\)

Hence, the maximum height the shell will attain is \(8.0{\rm{ km}}\).

The motion of the shell is represented as,

Figure: The motion of the shell

Since a triangle\(OAB\) is a right-angled triangle at\(A\). Applying Pythagoras theorem in the triangle\(OAB\).

\({\left( {{R_E} + x} \right)^2} = {R^2} + R_E^2\)

Here\({R_E}\) is the radius of the Earth,\(x\) is the distance between the horizontal line and the Earth’s surface, and\(R\) is the range of the projectile.

Rearranging the above equation to get an expression for the distance between the horizontal line and the Earth’s surface.

\(x = \sqrt {{R^2} + R_E^2} - {R_E}\)

Substitute\(32.0{\rm{ km}}\) for\(R\), and\(6.37 \times {10^3}{\rm{ km}}\) for\({R_E}\),

\(\begin{aligned}{}x = \sqrt {{{\left( {32.0{\rm{ km}}} \right)}^2} + {{\left( {6.37 \times {{10}^3}{\rm{ km}}} \right)}^2}} - \left( {6.37 \times {{10}^3}{\rm{ km}}} \right)\\ = \left( {0.080{\rm{ km}}} \right) \times \left( {\frac{{{{10}^3}{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\\ = 80.0{\rm{ m}}\end{aligned}\)

Hence, the Earth’s surface will be \(80.0{\rm{ m}}\) below the horizontal line parallel to the surface of the ship. This error will not play any significant role because \(80.0{\rm{ m}}\) is only \(1\% \) of the maximum height attained.