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

Question

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 $9.80 \frac{m}{s^{2}}$

Answer 1

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_{0 x}

) =0

\frac{m}{s}

Speed of egg

(v) = \sqrt{2 g h}

, where

g = 9.8 \frac{m}{s^{2}}

= \sqrt{2 (9.8) (46)} = 30.03 \frac{m}{s}

Let the egg drops on the persons

Answer 2

$S = μ t + \frac{1}{2} g t^{2}$

$\Rightarrow 44.2 = 0 t + \frac{1}{2} \times 98 \times t^{2}$

$\Rightarrow t^{2} = 3 s$

$∴$ Professor should be at distance of $3 \times 1.2 - 3.6$ m

Answer 3

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

$t = \frac{x_{p r o}}{v_{p r o}}$

$y - y_{0} = v_{0} t + \frac{1}{2} g t^{2}$

Substitute from the second equation into the second we get:

$44.2 = 0 + 4.9 \times \frac{x_{p r o}^{2}}{v_{p r o}^{2}}$

$x_{p r o}^{2} = \frac{{1.2}^{2} \times 44.2}{4.9} = 12.98$

$x_{p r o} = 3.6 m$

The professor should be 3.6m away from the building

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

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