A cannon ball is fired from ground level with a speed of 68.0 m/s at an angle of 34 deg. above the horizontal. A 13.0 m tall wall is located 125 m away from the cannon. a) By how much does the cannon ball clear the wall? b) what is the speed of the cannon ball when it is directly above the wall?
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time = 125 / [68 * cos 34] = 2.22s
vertical motion
s =ut + 0.5 at^2
s = 68 * sin 34 * 2.22 - 0.5 * 9.81 * 2.22^2
s = 60.24m
the cannon ball clears the wall = 60.24 - 13 = 47.24m
answer
v^2 =u^2 + 2as
v^2 = [68 sin 34]^2 - 2 * 9.81 * 60.24
v = 16.24
vertical component
horizontal component = 68 cos 34 = 56.37
speed of the cannon ball when it is directly above the wall
= sq root [ 16.24^2 + 56.37^2]
m/s
answer
vertical motion
s =ut + 0.5 at^2
s = 68 * sin 34 * 2.22 - 0.5 * 9.81 * 2.22^2
s = 60.24m
the cannon ball clears the wall = 60.24 - 13 = 47.24m
answer
v^2 =u^2 + 2as
v^2 = [68 sin 34]^2 - 2 * 9.81 * 60.24
v = 16.24
vertical component
horizontal component = 68 cos 34 = 56.37
speed of the cannon ball when it is directly above the wall
= sq root [ 16.24^2 + 56.37^2]
m/s
answer