Question #536ad

1 Answer
Oct 22, 2016

Answer:

#"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.