Answer:
The downward stone would hit the ground first but their speeds would be the same.
Explanation:
We will take downward direction be the positive side. Let v and -v be the speed of the downward and upward stone, respectively. Also g = 9.8 is the constant gravitational acceleration acting on both of them. We have the following equation of motions
Upward stone: [tex]s_u = -vt + gt^2/2[/tex]
Downward stone[tex]s_d = vt + gt^2/2[/tex]
The distance difference between them is
[tex]\Delta s = s_d - s_u = vt - (-vt) = 2vt[/tex]
Since v,t > 0, [tex]s_d - s_u >0[/tex], so [tex]s_d > s_u[/tex]. This means the downward stone will always be 'one-step-ahead' of the upward stone in the downward direction. Therefore, the downward stone would reach the ground first.
As for the speed, since they both start at the same height (same potential energy) with same speed (same kinetic energy), their total mechanical energy would be the same. So when then both hit the ground at different time but same height, their potential and kinetic energy would be the same, so are their speed.
