The equation of motion is h(t)=6+ut-½gt², where h(t) is the height above the water after t seconds, u=initial vertical speed of the diver, g=acceleration of gravity. When t=0, h(0)=6m.
Differentiate h(t):
dh/dt=u-gt=0 at maximum height, achieved when t=0.28s.
So, u-0.28g=0, and u=0.28g.
Therefore h(t)=6+0.28gt-½gt²=0 when the diver hits the water.
We have a quadratic in t which can be solved using the formula:
½gt²-0.28gt-6=0, t=(0.28gt±√(0.0784g²+12g))/g.
So t=0.28±√(0.0784+12/g).
We can substitute g=9.81 m/s² approx, so 12/g=1.2232 approx.
t=0.28+√1.3016=0.28+1.1409=1.42 seconds approx.
So it takes Eric 1.42 seconds to reach the water after leaving the diving board.