two stones

logistic_guy

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

x1=12gt2\displaystyle x_1 = \frac{1}{2}gt^2

The position of the second stone is:

x2=12g(t1.5)2\displaystyle x_2 = \frac{1}{2}g(t - 1.5)^2

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

v=g(t1.5)\displaystyle v = g(t - 1.5)

This gives:

t=vg+1.5\displaystyle t = \frac{v}{g} + 1.5

During this time, the first stone travels a distance:

x1=12gt2=12g(vg+1.5)2=12(9.8)(129.8+1.5)2=36.3719 m\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:

x2=12g(t1.5)2=12v2g=121229.8=7.3469 m\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:

Δx=x1x2=36.37197.3469=29.025 m\displaystyle \Delta x = x_1 - x_2 = 36.3719 - 7.3469 = \textcolor{blue}{29.025 \ \text{m}}
 
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