# You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along side the building at

Suman Cole 2021-03-07 Answered

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along side the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall.Take the free fall acceleration to be

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Bertha Stark
Height of building (h) = 46.0 m
Height of walling person (${h}_{1}$) = 1.2 $\frac{m}{s}$
Initial velocity of the egg is (${v}_{0x}$) =0 $\frac{m}{s}$
Speed of egg $\left(v\right)=\sqrt{2gh}$, where $g=9.8\frac{m}{{s}^{2}}$

Let the egg drops on the persons
###### Not exactly what you’re looking for?
Jeffrey Jordon

$S=\mu t+\frac{1}{2}g{t}^{2}$

$⇒44.2=0t+\frac{1}{2}×98×{t}^{2}$

$⇒{t}^{2}=3s$

$\therefore$ Professor should be at distance of $3×1.2-3.6$ m

###### Not exactly what you’re looking for?
Jeffrey Jordon

Take the positive y direction to be downward. When the egg fall it has to reach the professor height so, it has to travel $y-{y}_{0}=46-1.8=44.2m$. While the is falling the professor moves a distance x that we are interested in.

$t=\frac{{x}_{pro}}{{v}_{pro}}$

$y-{y}_{0}={v}_{0}t+\frac{1}{2}g{t}^{2}$

Substitute from the second equation into the second we get:

$44.2=0+4.9×\frac{{x}_{pro}^{2}}{{v}_{pro}^{2}}$

${x}_{pro}^{2}=\frac{{1.2}^{2}×44.2}{4.9}=12.98$

${x}_{pro}=3.6m$

The professor should be 3.6m away from the building