Question

A man was standing on top of an office building and ‘accidentally' dropped an rock off the roof How tall is the building if the total velocity is 123m/s and the horizontal velocity is 22m/s?

Answer 1

Answer:

The building is 746.666 meters high.

Explanation:

We know the total velocity of the rock ($v$), in meters per second, and its horizontal speed ($v_{x}$), in meters per second, at the moment right before its impact. It should be noted that the rock is experimenting a parabolic motion, which is the combination of a horizontal motion at constant speed and a free fall, which is an uniform accelerated motion due to gravity.

The final vertical speed of the rock ($v_{y}$), in meters per second, is determined by the following Pythagoric formula:

$v_{y} = \sqrt{v^{2}-v_{x}^{2}}$ (1)

If we know that $v = 123\,\frac{m}{s}$ and $v_{x} = 22\,\frac{m}{s}$, then the final vertical speed is:

$v_{y} = \sqrt{\left(123\,\frac{m}{s} \right)^{2}-\left(22\,\frac{m}{s} \right)^{2}}$

$v_{y} \approx 121.017\,\frac{m}{s}$

Now we determine the height of the building ($h$), in meters, by the use of the following kinematic expression:

$h = \frac{v_{y}^{2}-v_{y,o}^{2}}{2\cdot g}$ (2)

Where:

$v_{y,o}$ - Initial vertical speed of the rock, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

If we know that $v_{y,o} = 0\,\frac{m}{s}$, $v_{y} \approx 121.017\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$, then the height of the building is:

$h = \frac{v_{y}^{2}-v_{y,o}^{2}}{2\cdot g}$

$h = \frac{\left(121.017\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 746.666\,m$

The building is 746.666 meters high.