physics

An Olympic diver falls from rest from the platform 10m high. At what velocity does the diver hit the water is acceleration in 9.8m/s^2.
Do I assume "rest" means a initial velocity of 0? Could the formula be:
d= (vf^2-vi^2)/2(a) ?
as in d=10, a=9.8 vi=0
vf = √(d x 2(a)+ vi^2
vf= √(10 x 2(9.8) + 0^2
ans:v=14m/s
I'm not sure if this is right, the number seems too large... please help, thank you!

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  1. d = (1/2)(9.8)t^2
    10 = 4.9 t^2
    t = 1.43 seconds falling

    v = a t
    v = 9.8 (1.43) = 14 m/s , yes right
    Your way is fine. I just cheated to get it faster.

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