Asked by Amie
A stone is dropped from the roof of a high building. A second stone is dropped 1.51 later. How far apart are the stones when the second one has reached a speed of 12.6 ?
Answers
Answered by
Henry
Second Stone:
d = (V^2-Vo^2)/2g.
d = (158.76-0)/19.6 = 8.1 m.
Tf = (V-Vo)/g = (12.6-0)/9.8 = 1.29 s. =
Fall time.
First Stone:
Tf = 1.51 + 1.29 = 2.80 s.
d = Vo*t + 0.5g*t^2.
d = 0 + 4.9*2.8^2 = 38.3 m.
D = 38.3 - 8.1 = 30.2 Meters apart.
d = (V^2-Vo^2)/2g.
d = (158.76-0)/19.6 = 8.1 m.
Tf = (V-Vo)/g = (12.6-0)/9.8 = 1.29 s. =
Fall time.
First Stone:
Tf = 1.51 + 1.29 = 2.80 s.
d = Vo*t + 0.5g*t^2.
d = 0 + 4.9*2.8^2 = 38.3 m.
D = 38.3 - 8.1 = 30.2 Meters apart.
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.