A diver on the ledge of a diving platform jumos 2 meters int

A diver on the ledge of a diving platform jumos 2 meters into the air, then fall to the water below. The time for the entire dive was 3 seconds.

a) at what speed did the diver leave the platform?

b) how high above the ground was the platform?

c) what was the diver\'s impact velocity with the water?

please show all steps. hint:we should first find the time at the high tip then solve for others.

u = sqrt ( 2 *g * h)

u = sqrt ( 2 * 9.8 *2)

u = 6.26 m/s----------------------Answer------------

b) s = - U*t + 0.5 *g * t^2

S = - 6.26*3 + 0.5*9.8*9

S = 25.32 m---------Answer---------

v = sqrt(2 *g * (S+h))

v = sqrt (2*9.8*(27.32)

V = 23.14 m/s----------Answer------------