- #1

A stone is dropped into a river from a bridge 43.9m above the water. Another stone is thrown vertically down 1.00 s after the first is dropped. Both stones strike the water at the same time.

**What is the initial speed of the second stone?**

Well, I did this:

Y-Yo = 43.9m

a = -9.8 m/s^2

Vo1 = 0 m/s because it was just dropped

Y-Yo = Vot + 0.5at^2

43.9 = 0 + 0.5(9.8)t^2 = 4.9t^2 --> t1 = 2.99s

Since they both reached the water at the same time, t1 = 2.99s = t2.

Since stone 2 is thrown 1.00 s after stone 1, I decided to do: 2.99 + 1.00 = 3.99s to use for t in Y-Yo = Vot + 0.5at^2 for stone 2

So I got:

43.99 = Vo(3.99) + 0.5(-9.8)(3.99)^2 --> Vo = 30.5535 m/s

For some reason, I believe my method to be wrong because doesn't the t in Y-Yo = Vot + 0.5at^2 mean the total time it took to reach the water? And since they both reached the water at the same time, how can the t I used the second time be 3.99s? If I used 2.99s for t to get the Vo for stone 2, wouldn't that give me a Vo of 0? I am just confused as how to incorporate that 1.00s time lag and such :(

Also, because I've had much difficulty with these problems, can anyone give me some tips to solving these problems? Thanks so much!