The cannon on a battleship can fire a shell a maximum distance of 32.0 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 6.37×10³ km . How many meters lower will its surface be 32.0 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?

Solution:

Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}

we can say that \sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1. So, we have

\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}

We can solve for v₀ in terms of the other variables. That is

\displaystyle \text{v}_0=\sqrt{\text{gR}}

Substituting the given values, we have

\begin{align*}
\displaystyle \text{v}_0 & =\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)} \\
\displaystyle \text{v}_0 & =560.29\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}

The initial vertical velocity, v_0y, is calculated as

\begin{align*}
\text{v}_{\text{0y}} & =\text{v}_0\sin \theta _0 \\
& =\left(560.29\:\text{m/s}\right)\sin 45^{\circ}  \\
& =396.18\:\text{m/s}
\end{align*}

Therefore, the maximum height is

\begin{align*}
\text{h}_{\text{max}} & =\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\
\text{h}_{\text{max}} & =8000\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\

\end{align*}

Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

\begin{align*}
\text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(\text{R}+\text{d}\right)^2 \\
\left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(6.37\times 10^6+\text{d}\right)^2 \\
\text{d} & =\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6 \\
\text{d} & =80.37\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

This error is not significant because it is only about 1% of the maximum height computed in Part B.