You throw a small rock straight up from the edge of a highway bridge that crosses a river. the rock passes you on its way down, 5.00 s after it was thrown. what is the speed of the rock just before it reaches the water 25.0 m below the point where the rock left your hand? ignore air resistance.

Refer to the diagram shown below.
Note that air resistance is ignored, and the acceleration due to gravity is 9.8 m/s².

The vertical flight from A to B (the maximum height) and from B to C take equal amounts of time because the flight is parabolic.
The time of flight from A to C is given as 5 s, therefore the flight from A to B is 2.5 s.

Let u  = vertical launch velocity (m/s).
Because the vertical velocity at B is zero, therefore
u - (9.8 m/s²)*(2.5 s) = 0
u = 24.5 m/s

Let v =  the velocity with which the stone strikes the water.
The water surface is at a height of h = -25 m.
The equation governing the flight from A to the surface of the water is
v² = u² - 2gh
v² = (24.5 m/s)² - 2*(9.8 m/s²)*(-25 m)
    = 1090.25
v = 33.02 m/s


The velocity with which the stone strikes the water surface is 33.0 m/s (nearest tenth)

Then the answer would be d then