There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here
You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall. Â
Answer:
d=3.67 m
Explanation:
Height of building=46.0 m
First we need to find time taken by egg to reach 1.80 m above the surface
So to find time use below equation
[tex]S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s[/tex]
As velocity 1.20m/s is given and we have find time.So we can easily find the distance
So
[tex]distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m[/tex]