physics
posted by Anonymous .
A rock is dropped from rest off a cliff to the surface of a lake which is 6.2 m below. What is the velocity of the rock just as it hits the water?

9.8 m/s ?

using the formula vf^2=vo^2+ 2ad you can find out the answer. Because the stone is dropped the acceleration is 9.8m/s^2 and the vo=0m/s. The distance is aleady given to you. From here simply plug in the numbers and do the calculations to figure out vf. It should be around 11m/s