Asked by Alex
An artillery shell is fired horizontally from a cliff and 3 seconds later it hits the ground.
A)how high is the cliff?
B) if it’s initial velocity was 500m/s, how far does it land from the base of a cliff?
C) what was its speed as it hits the ground?
A)how high is the cliff?
B) if it’s initial velocity was 500m/s, how far does it land from the base of a cliff?
C) what was its speed as it hits the ground?
Answers
Answered by
oobleck
how far does it fall in 3 seconds?
s = 4.9t^2 = 4.9*9 = 44.1 m
it lands 500*3 = 1500 m from the cliff
Since the vertical speed is v = at = 3*9.81 = 29.43 m/s
the final speed is √(29.43^2 + 500^2) = 500.86 m/s
s = 4.9t^2 = 4.9*9 = 44.1 m
it lands 500*3 = 1500 m from the cliff
Since the vertical speed is v = at = 3*9.81 = 29.43 m/s
the final speed is √(29.43^2 + 500^2) = 500.86 m/s
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