Question

A cannonball, weighing 19.1 kg, is shot from a cannon sitting on
a ledge a height of 2.50 m above sea-level toward

a castle. It is shot with a velocity of 42.0 m/s at an angle of
55.0° above the horizontal. The cannonball impacts on the

castle wall 5.70 s later.

(a) At what height, in feet, does the cannonball hit the castle
wall?

(b) What is the cannonball’s impact velocity (magnitude and
direction) just before hitting the cliff?

(c) Using conservation of energy, determine the maximum height (in
m) reached by the cannonball along its trajectory?

(d) If the final velocity of the cannonball is 25.0 m/s, how much
work was done by air resistance?

Answer #1

a)

using 2nd equation of motion

h = 2.5 + 42* sin 55 * 5.7 - 0.5* 9.8* 5.7^2

h = 39.4 m = 139 ft

======

b)

horizontal velocity

vx = 42 cos 55 = 24.09 m/s

vertical velocity

vy = 42 sin x - 9.8* 5.7 = - 21.46 m/s

magnitude

v^2 = vx^2 + vy^2

v = 32.26 m/s

direction

x = arctan ( vy/vx) = 41.7 below horizontal

======

c)

using conservation of energy

0.5 m ( v sin x) ^2 = m g h

0.5* (42 sin 55)^2 = 9.8* h

h = 60.391 m

====

d)

W = change in kE = 0.5* 19.1* ( v^2 - 25^2) = - 3970 J

======

Comment before rate in case any doubt, will reply for sure.. goodluck

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