Question

The cannon on a battleship can fire a shell a maximum distance of 38.0 km.

(a) Calculate the initial velocity of the shell.
? m/s

(b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.)
? m

(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 38.0 km from the ship along a horizontal line parallel to the surface at the ship?
? m

Does your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)

a. The error could be significant compared to the size of a target.

b. The error is insignificant compared to the distance of travel.

c. The error is insignificant compared to the size of a target.

d. The error is significant compared to the distance of travel.

Answer 1

(a) for maximum range , $\theta$ = 45

so

v² = g * R

v = sqrt (gR)

v = sqrt (9.8 * 38000)

v = 610.24 m/s

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(b) Using kinematics in y-direction

at maximum height, velocity is zero

so,

h = v² / 2g

h = (610.24 * sin 45)² / 2 * 9.8

h = 9500 m

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(c) The radius of the earth is 6371 km

applying pythagoras, we have

h = sqrt ((d² + r² ) - r)

h = 6371.113 - 6371

h = 0.113325 Km

so,

h = 111.33 m