If we throw a body upwards at time t=0 with an initial speed s=25 m/s, what would the body's speed, position and velocity?

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²

Explanation:

This is a vertical throwing exercise.

The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.

Let's write the kinematics equations

        v = v₀ - g t

        v² = v₀² - g (y-y₀)

        y = y₀ + v₀ t - ½ g t²

in our case the initial velocity is

         v₀ = 25 m / s

and we zero the system at the launch point

        y₀ = 0

the equations remain

      v = 25 - 9.8 t

        v² = 25² - 9.8 y

        y = 0 + 25 t - ½ 9.8 t²

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²