Asked by sandhu
A football on the moon (1/6 earth's gravity) is kicked straight up with an initial velocity of +14.5 m/s. If it starts 2.7 m above the surface, how much time has passed until it is 16.8 m above the surface and on its way down?
Answers
Answered by
Damon
16.8 = 2.7 + 14.5 t -(1/12)9.8 t^2
or
14.1 = 14.5 t - .817 t^2
t^2 -14.5 t + 14.1 = 0
t = [ 14.5 +/- sqrt (14.5^2 -4*14.1) ]/2
t = [ 14.5 +/- 12.4 ] / 2
= 1.05 or 13.5
the 1.05 is on the way up
so the answer is 13.5 seconds
or
14.1 = 14.5 t - .817 t^2
t^2 -14.5 t + 14.1 = 0
t = [ 14.5 +/- sqrt (14.5^2 -4*14.1) ]/2
t = [ 14.5 +/- 12.4 ] / 2
= 1.05 or 13.5
the 1.05 is on the way up
so the answer is 13.5 seconds
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.