Asked by Robert
A ball is thrown in the -y direction off of a cliff with a velocity of 7m/s. If the ball takes 1.45s to reach the ground, how high off of the ground is the cliff?
Answers
Answered by
oobleck
y = h - 7t - 4.9t^2
so just solve for h in
h - 7(1.45) - 4.9*1.45^2 = 0
h = 20.45 m
not much of a cliff ...
so just solve for h in
h - 7(1.45) - 4.9*1.45^2 = 0
h = 20.45 m
not much of a cliff ...
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