Question

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 $6.37 \times 10^3 \textrm{ km}$. 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?

a) $560 \textrm{ m/s}$

b) $8.00 \textrm{ km}$

c) The Earth's surface will be $80 \textrm{ m}$ below the horizontal line parallel to the surface at the ship. The assumption of a flat Earth is acceptable for a range of $32 \textrm{ km}$ since this discrepancy represents only $1%$ of the maximum height of the projectile.

# OpenStax College Physics, Chapter 3, Problem 33 (Problems & Exercises)  In order to watch this solution you need to have a subscription.

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• • Video Transcript
Hello lesiramdelgado, I tried to explain this at 0:56 in the video. Given the range formula $R = \dfrac{v_o^2 \sin{2\theta}}{g}$ we know $R$ will be at its maximum when $\sin{2\theta}$ is at its maximum. The largest possible value for $\sin({\textrm{any number at all}})$ is 1. This will happen when taking the sine of 90. Since the angle is getting multiplied by 2 in the formula, we'll be taking the sine of 90 when the angle is $45^\circ$.