Question

A ball is dropped from a height of 10m. At the same time, another ball is thrown

Answer 1

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

$y_1 = 10 - \frac{1}{2}gt^2$ (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 $y_2,$ is given by

$y_2 = v_0t - \frac{1}{2}gt^2$ (2)

At the instant the two balls collide, they will have the same displacement, therefore

$y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2$

or

$v_0t = 10\:\text{m}$

Solving for t, we get

$t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}$

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

$y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2$

$\:\:\:\:\:\:\:= 5.1\:\text{m}$