Asked by Myke
Standing on a bridge, you throw a 3.0 g stone straight upward with speed of 2.5 m/s. How long does it take to return to the original position?
Answers
Answered by
Steve
while rising,
v = v0 + at
0 = 2.5 - 9.8t
t = .255
That's the time to stop rising.
Total time in the air is thus 0.510 sec
v = v0 + at
0 = 2.5 - 9.8t
t = .255
That's the time to stop rising.
Total time in the air is thus 0.510 sec
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