Kinetic Energy

[songokou77]

A 66.0 kg diver is 4.90 m above the water, falling at speed of 8.20 m/s. Calculate her kinetic energy as she hits the water. (Neglect air friction)

Then the problem hints:Use conservation of mechanical energy.

So i'm thinking that conservation is K_f+U_f=K_i+U_i but there must something wrong with my calculations>
K=1/2(m*v^2) and U=mgh right?

[stunner5000pt]

ok let's say that delta (the triangle) is represented by d in my calcultions


dK + dU = 0 conservation of energy

KE (final) - KE (Initial) + PE (Initial) - PE (final) = 0

sinceu want a value for KE final

PE (final) = m g (0) = 0 becuase as diver hits water height is zero

KE final = PE (initial) + KE (initial)

now sub in and see waht you get

[songokou77]

From KE_final=PE(initial)+KE(initial) i get:
(66.0)(9.8)(4.90)+1/2(66.0)(8.20^2)= 5388.2 and it was right thanxs a lot