Sam and Terry find a 29.0 m deep wishing well and decide to make a wish. Terry throws a penny down the well at 9.00 m/s and Sam throws a penny up in the air at the same speed. If they threw them at the same time, how much time would there be between splashes?

Answer 1

Terry:
d = Vo*t + 0.5g*t^2 = 29.
9t + 4.9t^2 = 29
4.9t^2 + 9t -29 = 0.
Use Quadratic Formula:
t = -9 +- Sqrt(9^2+4*4.9*29)/9.8 = 1.68 s. = Time to reach bottom of well.

Sam:
V = Vo + g*Tr = 0.
9 - 9.8Tr = 0
Tr = 0.918s = Rise time.
Tf1 = Tr = 0.918s = Fall time to launching point.

d = Vo*Tf2 + 0.5g*TF2 = 29.
9Tf2 + 4.9(Tf2)^2-29 = 0
Tf2 = 1.68 s.
T = Tr+Tf1+Tf2 = 0.918 + 0.918 + 1.68 = 3.52 s.

T-t = 3.52 - 1.68 = 1.84 s. Between splashes.