# Question #536ad

##### 1 Answer

#### Answer:

#### 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

More specifically, you can use the fact that the balls will travel for the *same period of time*, let's say

So, the first ball is dropped from a height of

#h_1 = v_(01) * t + 1/2 * g * 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 = 1/2 * g * t^2#

Now, if the first ball travels a distance

#h_"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_"collision" = v_(02) * t - 1/2 * g * t^2#

Now all you have to do is use equation

#15 - color(red)(cancel(color(black)(1/2 * g * t^2))) = v_(02) * t - color(red)(cancel(color(black)(1/2 * g * 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

#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**.