Question

You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongside the building at a constant speed of 1.20m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg?

Answer 1

3.6m

Step-by-step 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:

`h=v_(0)t+(1)/(2)gt^2`

where
`h` is the height,
`v_(0)` is the initial velocity, in this case
`v_(0)=0`.
`g` is the acceleration of gravity:
`g=9.81m/s` and
`t` is time, thus:

`h=(1)/(2)gt^2`

clearing for time:

`2h=gt^2\\(2h)/(g)=t^2\\\sqrt{(2h)/(g)} =t`

we know that the egg has to fall for 44.2m, so
`h=44.2`, and
`g=9.81m/s`, so we the time is:

`t=\sqrt{(2(44.2m))/(9.8m/s^2) }=\sqrt{(88.4m)/(9.81m/s^2) } =√(9.011s^2)= 3.002s`

Finally, if the professor has a speed of
`v=1.2m/s`, it has to be at a distance:

`d=vt`

and t=3.002s:

`d=(1.2m/s)(3.002s)=3.6m`

so the answer is the professor has to be 3.6m far from the building when you release the egg