A diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s. based on speed how high above the board will the diver reach?

When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula

v^2 = u^2 - 2gh

where

v = 0 (velocity at maximum height)

u = 1.2 m/s, intial upward velocity

g = - 9.8 m/s^2, gravitational acceleration (downward)

h = maximum height attained above the diving board.

Therefore

0 = 1.2^2 - 2*9.8*h

h = 1.2^2 / (2*9.8) = 0.0735 m

Answer: 0.074 m (nearest thousandth)