Question

At t = 0, Ball 1 is dropped from the top of a 22 m-high building. At the same instant Ball 2 is thrown straight up from the base of the building with an initial velocity of +12 m/s. Assume that Ball 1 starts from rest and that air friction can be ignored. At what height will the two balls pass each other? What is the velocity of each ball at the moment they are at the same height?

Answer 1

Answer:

The two balls pass each other at a height of 5.53 m

vf1=17.97 m/s

vf2=-5.96 m/s

Explanation:

Vertical Motion

An object thrown from the ground at speed vo, is at a height y given by:

$y=vo.t-g.t^2/2$

Where t is the time and $g=9.8\ m/s^2$

Furthermore, an object dropped from a certain height h will fall a distance y, given by:

$y=g.t^2/2$

Thus, the height of this object above the ground is:

$H = h-g.t^2/2$

The question describes that ball 1 is dropped from a height of h=22 m. At the same time, ball 2 is thrown straight up with vo=12 m/s.

We want to find at what height both balls coincide. We'll do it by finding the time when it happens. We have written the equations for the height of both balls, we only have to equate them:

$vo.t-g.t^2/2=h-g.t^2/2$

Simplifying:

$vo.t=h$

Solving for t:

$t=h/vo=22/12=1.833\ s$

The height of ball 1 is:

$H = 22-9.8.(1.833)^2/2$

H = 5.53 m

The height of ball 2 is:

$y=12\cdot(1.833)-9.8\cdot(1.833)^2/2$

y=5.53 m

As required, both heights are the same.

The speed of the first ball is:

$vf1=g.t=9.8\cdot 1.833=17.97\ m/s$

vf1=17.97 m/s

The speed of the second ball is:

$vf2=vo-gt=12-9.8\cdot 1.833=-5.96\ m/s$

vf2=-5.96 m/s

This means the second ball is returning to the ground when both balls meet