Question

Question #536ad

Answer 1

`"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_"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 `"15 m"` at `t=0`. The distance covered, let's say `h_1`, will be

`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_1`, then it will be located at a height

`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"`, in the same time `t`

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

Now all you have to do is use equation `color(orange)("(*)")` to find the time `t`

`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 `color(orange)("(*)")` to find the value of `h_"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.