A child on a bridge throws a rock straight down to the water below. The point where the child released the rock is 74 m above the water and it took 2.7 s for the rock to reach the water. Determine the rock's velocity (magnitude & direction) at the moment the child released it. Also determine the rock's velocity (magnitude & direction) at the moment it reached the water. Ignore air drag.

The rock's altitude y at time t, thrown with initial velocity v, is given by

where is the acceleration due to gravity.

After t = 2.7 s, the rock reaches the water (0 altitude), so

so the rock was thrown with a velocity with magnitude 14 m/s and downward direction.

Its velocity at time t is (with no horizontal component), so that at the moment it hits the water, its velocity is

That is, its final velocity has an approximate magnitude of 41 m/s, also directed downward.

he . if i may ask what i your question exactly?

explanation:

the body's situation won't change the forces are equal opposit and on the same body