I seem to have some irregularities with a specific problem which goes like this:(adsbygoogle = window.adsbygoogle || []).push({});

"A stone is thrown upward from the edge of a cliff, with a speed of 70m/s. If, when it is 120m above the hand on its return journey, a second stone is dropped over the edge of the cliff, when and where will the stones be together?"

A nice little illustration:

http://img381.imageshack.us/img381/8154/testing6dv.th.jpg [Broken]

A graph to illustrate the velocities of the 2 stones:

http://img403.imageshack.us/img403/3606/testing6cg.th.jpg [Broken]

Now first, what I did was calculate the final velocity of stone #1:

2a + Δd = Vf^2 - Vi^2

2(10m/s^2) + (Δd) = Vf^2 - (-70m/s)^2

Vf^2 = 2500

Vf = 50m/s

Using this final velocity, I found the distance of stone #2:

2a + Δd = Vf^2 - Vi^2

2(10m/s^2) + (Δd) = (50m/s)^2 - (0m/s)^2

Δd = 2480m

Now, the only time in which the stones can be together is when their final velocities are the same. Knowing this, we use this equation:

Δd = (vi)(Δt)+1/2a(Δt)^2

(2480m + 120m) = (50m/s)(Δt) + 1/2(10m/s)(Δt)^2 (In this step, the total distance is the distance from stone 2 but also added from stone 1)

52/5 = Δt^3

Δt = 2.2

The time in which the two stones meet is 2.2s but the answer is 2.4.

Did I do something wrong in my process, or is my approach completely wrong?

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# Homework Help: Motion Word Problems Help!

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