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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.