A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 m/s. The cliff is 52 m above the water's surface. How long does it take for the stone to fall to the water? With what speed does it strike the water?
The Attempt at a Solution
So far, I was able to figure out the time:
dy= 1/2gt^2 where g=9.81 m/s^2
t= 3.26 seconds
I thought it would be correct to use Vy=gt to find the speed at which the stone strikes the water, therefore being (9.81 m/s^2)(3.26 s)= 31.98 m/s. But when i checked the answer key and talked to the teacher, she said that the answer was 36.889 m/s, and I was wrong.
So can someone tell me what I'm doing wrong and how to find the speed at which the stone strikes the water?