Question

You are on the roof of the physics building 46.0 meters above theground. Your physics professor, who is 1.8 meters tall, is walking alongside the building at a constant speed of 1.20 meters per second. If you wish to drop an egg on your professor’s head, where should the professor be when v=you release the egg? Assume the egg is in free fall with no losses due to friction or air resistance.

Answer 1

Answer:

The professor should be $d=3.6m$ away of the impact point, walking towards it.

Explanation:

We will calculate how much time an egg takes to fall from the initial position to the final position, and then use that time to calculate how much distance the professor moved.

We use the equation for accelerated motion (on the vertical direction):

$y_f=y_i+v_{0y}t+\frac{a_yt^2}{2}$

Since it departs from rest and the acceleration is that of gravity, we have:

$y_f=y_i+\frac{gt^2}{2}$

Which means:

$t=\sqrt{\frac{2(y_f-y_i)}{g}}$

Having the ground as reference and taking the upwards direction as positive, for our values we have:

$t=\sqrt{\frac{2(1.8m-46m)}{-9.8m/s^2}}=3s$

And in that time the professor moved $d=vt=(1.2m/s)(3s)=3.6m$