The cannon on a battleship can fire a shell a maximum distance of 44.0 km. (a) C

Question

The cannon on a battleship can fire a shell a maximum distance of 44.0 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 44.0 km from the ship along a horizontal line parallel to the surface at the ship?

m

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

The error is significant compared to the distance of travel.

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

  The error is insignificant compared to the distance of travel.

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

Explanation / Answer

(a) The cannon's range depends upon the angle of the barrel (known as elevation).
I'm going to assume an elevation of = 45° since tan 45° = 1. This means that
for each meter of height the shell rises, it also moves downrange a meter of
horizontal distance.

R = horizontal range = 44.0 km = 44,000 m
= angle of the barrel = 45 °
g = gravitational acceleration = 9.81 m/s²
v = initial velocity of the cannon shell = to be determined

R = (v²sin 2)/g
Rg = v²sin 2
Rg/sin 2 = v²
v² = Rg/sin 2
v = (Rg/sin 2)
v = [(44,000 m)(9.81 m/s²)/sin 2(45°)]
v = 657 m/s

(b) H = maximum height reached by the cannon shell = to be determined

H = (v²sin² )/2g
H = [(657 m/s)²sin² 45°]/2(9.81 m/s²)
H = 11,000 m

(c) r = radius of the earth = 6,371 km = 6.37E+06 m

Average curvature of the earth = 7.98 in / 1 mi
Average curvature of the earth = 20.2692 cm / 1.6093 km

Average curvature of the earth = 12.6 cm/km

So, over a distance of 44.0 km the earth surface drops
44.0 km x 12.6 cm/km = 554.4 cm = 5.54 m

d) The error is insignificant compared to the distance of travel.
The error is insignificant compared to the size of a target.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.