Question

You are on the roof of the physics building of your school, 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, assuming that the egg encounters no appreciable air drag? Express your answer in meters to three significant figures.

Answer 1

The professor should be 3.60 meters away from the place where the egg would fall.

Step-by-step explanation:

To calculate this first we need to calculate the time it would take for the egg to fall the distance from the roof to the professor's head (46m-1.8m=44.2m).

The formula for free fall is:

y=1/2*g*
`t^(2)`

solving for t it would be

`\sqrt{(y)/(g*1/2)} =t`

replacing

`\sqrt{(44.2m)/(9.8m/s^2*1/2)} = t = 3.00s`

Now that we have the time it would take the egg to get from the roof to the desired height we need to calculate the distance the professor would travel in that time so we know to release the egg when he's exactly at that distance.

Since he is traveling at a constant speed the formula is

x=v*t

replacing

1.20m/s * 3.00 s = x = 3.60 m