Answer
The professor should be 3.60 meters away from the target point at the moment the egg is released.
Work Step by Step
The vertical distance from the top of the building to the professor's head is 44.2 m. We can find the time $t$ for the egg to fall this distance.
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}} = \sqrt{\frac{(2)(44.2~m)}{9.80~m/s^2}} = 3.00~s$
We can find the distance the professor moves in 3.00 seconds.
$x = vt = (1.20~m/s)(3.00~s) = 3.60~m$
The professor should be 3.60 meters away from the target point at the moment the egg is released.
