Asked by cory
A stone is thrown vertically upward with a speed of 11.0 m/s from the edge of a cliff 80.0 m high. How much later does it reach the bottom of the cliff?
Answers
Answered by
macfield musonda
v=u+at
0=11.0m/s+(-9.81)t
11.0/9.81=t
t=1.121304791
total time
t=2t
t=2*1.121304791
t=2.242609582
t=2.24s
0=11.0m/s+(-9.81)t
11.0/9.81=t
t=1.121304791
total time
t=2t
t=2*1.121304791
t=2.242609582
t=2.24s
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