Asked by Amy
The captain sinks a ship located x distance away with a cannon located on the top of a cliff that is 65m high. if the cannon ball has a velocity of 55m/s and the captain sinks the ship, how far was the ship from the base of the cliff.
I got 731.5 meters. is that right?
I got 731.5 meters. is that right?
Answers
Answered by
MathMate
ASSUMING the cannon direction was horizontal, and neglect air resistance.
Time to free-fall 65m, t:
65 = ut+(1/2)gt²
u=0 (initial vertical velocity)
t=√(65*2/9.81)=3.64 s.
Horizontal distance = 3.64*ux
= 3.64*55
= 200.2m
Can you explain how you got 731.5m?
Time to free-fall 65m, t:
65 = ut+(1/2)gt²
u=0 (initial vertical velocity)
t=√(65*2/9.81)=3.64 s.
Horizontal distance = 3.64*ux
= 3.64*55
= 200.2m
Can you explain how you got 731.5m?
Answered by
bobpursley
I do not know in what direction the cannon was aimed, particularly, the elevation.
Assuming horizontal, and no one would shoot a cannon horizontal, but
time to fall 65m...
65=1/2 g t^2 or
t= sqrt 130/9.8= 3.64sec
distance horizontal= 55m/s*3.64 which is not your answer.
Assuming horizontal, and no one would shoot a cannon horizontal, but
time to fall 65m...
65=1/2 g t^2 or
t= sqrt 130/9.8= 3.64sec
distance horizontal= 55m/s*3.64 which is not your answer.
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