Question

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

vertically upwards at an initial speed of 10m/sec. How high above the ground will the two balls
collide?

Answer 1

5.1 m

Step-by-step 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 - (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 - (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 - (1)/(2)gt^2 = v_0t - (1)/(2)gt^2`

or

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

Solving for t, we get

`t = \frac{10\:\text{m}}{v_0} = \frac{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} - (1)/(2)(9.8\:\text{m/s}^2)(1\:\text{s})^2`

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