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petitsreves

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**Homework Statement**

Rock A is thrown straight up from the top of a 25m tall building at a speed of 20m/s. Exactly 1 second later Rock B is thrown in such a way that it lands on the ground at the base of the building at the same instant that Rock A lands. What is the initial velocity for Rock B? (Air friction is neglected)

Answer: 14m/s [up]

ΔtA - 1 = ΔtB

ViA = 20m/s [up]

a = 9.8m/s^2 [down]

Δd(building) = 25m

ViB = ?

**My attempt:**

Rock A going up: vf^2 = vi^2 + 2aΔd

0^2 - 20^2 = 2(-9.8)Δd

Δd = 20.408m

t = (2Δd)/(vf + vi)

= (2 x 20.408)/20

t = 2.04s

Rock A going down: Δd = 20.408 + 25 = 45.408m

Δd = viΔt + 0.5a(Δt)^2

45.408 = 0.5 (-9.8)(Δt)^2

Δt = 3.04s

Rock A going up and then down: Δt = 2.04 + 3.04 = 5.08s

Rock B going up and then down: 4.08s (5.08 - 1 because of the 1s delay in throwing)

Then from here I'm stuck. I know the total time Rock B takes to go up and down is 4.08s. I know the building is 25m tall. I know acceleration is 9.8m/s^2 [down]. However, I don't know how those 4.08s are split between going up and down, nor do I know the distance that Rock B travels up before falling down. I already tried calculating Rock B's initial velocity if it was dropped instead of thrown (question doesn't specify, only "thrown in such a way"), but the answer isn't right.

I also attempted to substitute.

Rock B going up: t1 = Δv/a = vi/a

Rock B going down: t2 = √(vi^2 + 50 x 9.8 / 9.8^2)

t1 + t2 = 4.08s and substitute the above to solve for vi, but that didn't come out to the right answer either.

Any hints or help on how to work with what I have would be greatly appreciated.