Let's consider the motion of the two balls separately to determine the point at which they collide.
For the ball that is dropped, we can use the equation for vertical motion:
y = u * t + (1/2) * g * t^2
where: y is the displacement (height), u is the initial vertical velocity (0 m/s for the dropped ball), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
For the ball that is thrown vertically upward, we can use the same equation, but with a different initial velocity:
y = v * t + (1/2) * g * t^2
where: v is the initial vertical velocity (10 m/s for the thrown ball).
Since the two balls collide, their displacements (y) will be the same when they meet. Let's assume the time of collision is t_collide.
For the dropped ball: y = 0 * t_collide + (1/2) * g * t_collide^2
For the thrown ball: y = 10 * t_collide + (1/2) * g * t_collide^2
Setting these two equations equal to each other:
0 * t_collide + (1/2) * g * t_collide^2 = 10 * t_collide + (1/2) * g * t_collide^2
Simplifying:
10 * t_collide = 0
Since the time cannot be zero, we conclude that the two balls will never collide in mid-air.