Asked by guy
On Earth a rock falls an unknown vertical distance from a resting position and lands in a lake. If it takes the rock 2.5 seconds to fall, how high is the cliff that the rock fell from?
Answers
Answered by
jai
the formula for vertical distance is,
d = (v0)*t-(1/2)g*t^2
where v0=initial velocity, t=time, and g=acceleration due to gravity, which is approximately 9.8 m/(s^2)
*now since this is freefall, v0 = 0 and the formula reduces to:
d = -(1/2)g*t^2
d = -(1/2)*9.8*(2.5^2)
then solve for d. units in m. (get its absolute value since it's distance)
d = (v0)*t-(1/2)g*t^2
where v0=initial velocity, t=time, and g=acceleration due to gravity, which is approximately 9.8 m/(s^2)
*now since this is freefall, v0 = 0 and the formula reduces to:
d = -(1/2)g*t^2
d = -(1/2)*9.8*(2.5^2)
then solve for d. units in m. (get its absolute value since it's distance)
Answered by
guy
THANK YOU =0]
Answered by
Anonymous
On Earth a rock falls an unknown vertical distance from a resting position and lands in a lake. If it takes the rock 2.5 seconds to fall, how high is the cliff that the rock fell from?
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.