Question #536ad

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Question #536ad

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Answer 1 · 2016-10-22T00:44:07

$11.6 m$ $"11.6 m"$

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 $h_{collision}$ $h_"collision"$ .

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

So, the first ball is dropped from a height of $15 m$ $"15 m"$ at $t = 0$ $t=0$ . The distance covered, let's say $h_{1}$ $h_1$ , will be

$h_{1} = v_{01} \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $h_1 = v_(01) * t + 1/2 * g * t^2$

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

$h_{1} = \frac{1}{2} \cdot g \cdot t^{2}$ $h_1 = 1/2 * g * t^2$

Now, if the first ball travels a distance $h_{1}$ $h_1$ , then it will be located at a height

$h_{collision} = 15 - h_{1}$ $h_"collision" = 15 - h_1$

$h_{collision} = 15 - \frac{1}{2} \cdot g \cdot t^{2} (*)$ $h_"collision" = 15 - 1/2 * g * t^2" " " "color(orange)("(*)")$

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

$h_{collision} = v_{02} \cdot t - \frac{1}{2} \cdot g \cdot t^{2}$ $h_"collision" = v_(02) * t - 1/2 * g * t^2$

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

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

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.

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Question #536ad

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