You are at standing at a window in the University tower that is 42.0 meter above the ground. Your physics professor (who is 2.12 m tall) is walking on the ground below, approaching to tower at a speed of 1.12 m/s. You are holding an egg, which you would like to drop on your professor's head. If you drop the egg from rest, how far in advance of the tower should your professor be (in distance) at the instant of release such that the egg hits him directly on the head as he passes below the window?
How long to drop 42 - 2.12 = 39.88 meters?
h = Hi + Vi t + (1/2) a t^2
0 = 39.88 + 0 - 4.9 t^2
t^2 = 39.88/4.9
so
t = 2.85 seconds
how far does he walk in 2.85 seconds?
d = 1.12 * 2.85 = 3.2 meters
To calculate the distance your professor should be in advance of the tower when you drop the egg, we need to consider the time it takes for the egg to fall and the distance your professor walks during that time.
First, let's calculate the time it takes for the egg to fall from the window to the ground. We can use the equation for free-fall motion:
h = 1/2 * g * t^2
where h is the height, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2). We can rearrange the equation to solve for t:
t = √(2h / g)
Given that the height of the window above the ground is 42.0 meters, the time it takes for the egg to fall is:
t = √(2 * 42.0 / 9.8) = √8.5714 ≈ 2.93 seconds
Now, we need to calculate the distance your professor walks during this time. The distance your professor walks is equal to his velocity multiplied by the time it takes for the egg to fall:
Distance = Velocity * Time
Given that your professor's velocity is 1.12 m/s and the time is 2.93 seconds, we can calculate the distance as:
Distance = 1.12 m/s * 2.93 s ≈ 3.28 meters
Therefore, your professor should be approximately 3.28 meters in advance of the tower when you drop the egg such that it hits him directly on the head as he passes below the window.