Answer:
3.6m
Explanation:
if you are at a building that is 46m above the ground, and the professor is 1.80m, the egg must fall:
46m - 1.80m = 44.2m
the egg must fall for 44.2m to land on the head of the professor.
Now, how many time this takes?
we have to use the following free fall equation:
[tex]h=v_{0}t+\frac{1}{2}gt^2[/tex]
where [tex]h[/tex] is the height, [tex]v_{0}[/tex] is the initial velocity, in this case [tex]v_{0}=0[/tex]. [tex]g[/tex] is the acceleration of gravity: [tex]g=9.81m/s[/tex] and [tex]t[/tex] is time, thus:
[tex]h=\frac{1}{2}gt^2[/tex]
clearing for time:
[tex]2h=gt^2\\\frac{2h}{g}=t^2\\\sqrt{\frac{2h}{g}} =t[/tex]
we know that the egg has to fall for 44.2m, so [tex]h=44.2[/tex], and [tex]g=9.81m/s[/tex], so we the time is:
[tex]t=\sqrt{\frac{2(44.2m)}{9.8m/s^2} }=\sqrt{\frac{88.4m}{9.81m/s^2} } =\sqrt{9.011s^2}= 3.002s[/tex]
Finally, if the professor has a speed of [tex]v=1.2m/s[/tex], it has to be at a distance:
[tex]d=vt[/tex]
and t=3.002s:
[tex]d=(1.2m/s)(3.002s)=3.6m[/tex]
so the answer is the professor has to be 3.6m far from the building when you release the egg