two stones

[logistic_guy]

A stone is dropped from the roof of a high building. A second stone is dropped \(\displaystyle 1.50 \ \text{s}\) later. How far apart are the stones when the second one has reached a speed of \(\displaystyle 12.0 \ \text{m/s}\)?

 

[khansaheb]

A stone is dropped from the roof of a high building. A second stone is dropped \(\displaystyle 1.50 \ \text{s}\) later. How far apart are the stones when the second one has reached a speed of \(\displaystyle 12.0 \ \text{m/s}\)?

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[logistic_guy]

The position of the first stone is:

\(\displaystyle x_1 = \frac{1}{2}gt^2\)

The position of the second stone is:

\(\displaystyle x_2 = \frac{1}{2}g(t - 1.5)^2\)

We can use the velocity (speed) equation to calculate the time:

\(\displaystyle v = g(t - 1.5)\)

This gives:

\(\displaystyle t = \frac{v}{g} + 1.5\)

During this time, the first stone travels a distance:

\(\displaystyle x_1 = \frac{1}{2}gt^2 = \frac{1}{2}g\left(\frac{v}{g} + 1.5\right)^2 = \frac{1}{2}(9.8)\left(\frac{12}{9.8} + 1.5\right)^2 = 36.3719 \ \text{m}\)

The second stone travels a distance:

\(\displaystyle x_2 = \frac{1}{2}g(t - 1.5)^2 = \frac{1}{2}\frac{v^2}{g} = \frac{1}{2}\frac{12^2}{9.8} = 7.3469 \ \text{m}\)

The distance between the two stones is:

\(\displaystyle \Delta x = x_1 - x_2 = 36.3719 - 7.3469 = \textcolor{blue}{29.025 \ \text{m}}\)