# Question 536ad

Oct 22, 2016

$\text{11.6 m}$

#### Explanation:

The key to this problem is the time needed for both of the balls to reach the height at which they collide, let's say ${h}_{\text{collision}}$.

More specifically, you can use the fact that the balls will travel for the same period of time, let's say $t$, before colliding.

So, the first ball is dropped from a height of $\text{15 m}$ at $t = 0$. The distance covered, let's say ${h}_{1}$, will be

${h}_{1} = {v}_{01} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

Since the ball is being dropped and not thrown, you can say that its initial velocity will be equal to zero, and so

${h}_{1} = \frac{1}{2} \cdot g \cdot {t}^{2}$

Now, if the first ball travels a distance ${h}_{1}$, then it will be located at a height

${h}_{\text{collision}} = 15 - {h}_{1}$

h_"collision" = 15 - 1/2 * g * t^2" " " "color(orange)("(*)")

from the ground when it collides with the second ball. This means that the second ball must travel the same distance, i.e. ${h}_{\text{collision}}$, in the same time $t$

${h}_{\text{collision}} = {v}_{02} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$

Now all you have to do is use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to find the time $t$

$15 - \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{2} \cdot g \cdot {t}^{2}}}} = {v}_{02} \cdot t - \textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{2} \cdot g \cdot {t}^{2}}}}$

This gets you

15 = v_(02) * t implies t = (15 color(red)(cancel(color(black)("m"))))/(18color(red)(cancel(color(black)("m")))"s"^(-1)) = "0.833 s"

You can now plug this into equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to find the value of ${h}_{\text{collision}}$

h_"collision" = "15 m" - 1/2 * "9.81 m" color(red)(cancel(color(black)("s"^(-2)))) * 0.833^2 color(red)(cancel(color(black)("s"^2)))

h_"collision" = color(green)(bar(ul(|color(white)(a/a)color(black)("11.6 m")color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.