Asked by Anonymous
A ball is thrown horizontally of a cliff. If the initial speed of the ball is 17 m/s and the cliff is 77 m high, how far from the base of the cliff will the ball land in the water below? Calculate the answer in meters (m) and rounded to three significant figures.
Answers
Answered by
success
S=at/2.
77=5t^2
t=15.4s
range=vt
R=261.8m
77=5t^2
t=15.4s
range=vt
R=261.8m
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