5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
[tex]y_1 = 10 - \frac{1}{2}gt^2[/tex] (1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground [tex]y_2,[/tex] is given by
[tex]y_2 = v_0t - \frac{1}{2}gt^2[/tex] (2)
At the instant the two balls collide, they will have the same displacement, therefore
[tex]y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2[/tex]
or
[tex]v_0t = 10\:\text{m}[/tex]
Solving for t, we get
[tex]t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}[/tex]
We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):
[tex]y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2[/tex]
[tex]\:\:\:\:\:\:\:= 5.1\:\text{m}[/tex]
