A rock is dropped into a pool from the end of a board 5.20m above the surface of

Question

A rock is dropped into a pool from the end of a board 5.20m
above the surface of the pool. This rock hits the water with a certain velocity and then sinks to the bottom of the pool at this same constant velocity. It reaches the bottom 4.80sec after it was dropped.
A) How deep is the pool?
B) Suppose that all the water evaporates from the pool. The rock is now thrown
from the board so that it again reaches the bottom in 4.80sec. What
initial velocity would it need to do this?

I did:

Vx^2 = Vxo^2 + 2ax (x - xo)

Vx^2 = 0 + 2(9.8)(5.2-0)

Vx^2 = 101.92

Vx = 10.1 m/s


Then I found how long it takes for the rock to just hit the water.

5.2 = 0 + 0 + (.5)(9.8)(t^2)
t = 1.03 s

So now distance = velocity x time

D = (10.1)(4.8-1.03) = 38.077m <--- depth of pool

Now is where I am somewhat confused.

38.077+5.2 = 43.258 total distance

43.258 = 0 + Vox(4.8) + (4.9)(4.8^2)

I get the initial velocity needed as 32.5 m/s, is this correct?

Explanation / Answer

I got Vox = -14.51 m/s. Notice the negative sign which signifies that the initial velocity was upwards.

The equations you have used are absolutely correct. I think you messed in the final step. Think about it this way: Initially there was some resistance to the rock because of the water, which nullified the acceleration due to gravity 'g' and it reached the bottom of the pool in 4.8 seconds. Now, in the absence of the resistance, it will experience 'g' all the way to the bottom; so without any initial velocity, it should reach the bottom quicker than 4.8 seconds. So, if it has to reach the bottom in 4.8 seconds the initial velocity has to be upwards.

Hope this helps. Good luck!

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