Question

The cannon on a battleship can fire a shell a maximum distance of 50.0 km.

(a) Calculate the initial velocity of the shell.


(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.)


(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 50.0 km from the ship along a horizontal line parallel to the surface at the ship?

Answer 1

a)

for maximum horizontal distance ,

$\theta$ = angle of launch = 45

X = maximum horizontal distance = 50 km = 50000 m

v = speed of launch

maximum horizontal distance traveled is given as

X = v² Sin2$\theta$ /g

50000 = v² Sin(2 x 45) /9.8

v = 700 m/s

b)

maximum height is given as

H_max = v² Sin²$\theta$ /(2g)

H_max = 700² Sin²45 /(2 x 9.8)

H_max = 12500 m