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.