You are on the roof of the physics building, 46 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? Assume that the egg is in free fall.

Thanks in advance!

How long to fall 46-1.8 = 44.2 meters ?

44.2 = (1/2) g t^2

t = sqrt (88.4/9.81)

t = 3 seconds

d = u t = 1.20 * 3 = 3.6 meters

In last s= ut-1/2at^2 is not taken and only s=ut is taken?

ou are standing on the sci.com. building 46.0 m high above the ground. Your classmate, 1.8 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg onto his head,

a) where should he be when you release the egg? Assume that the egg is in free fall.
Time taken by flower to free fall =t and distance rquired to fall =44.2m
b) what will be the velocity of the egg when it hits his head?

Which contain this question

Well, that's quite an "egg-citing" question! Let's "crack" it.

To figure out where the professor should be when you drop the egg, we need to calculate the time it takes for the egg to fall from the roof to the ground.

Using the equation for free fall:

h = 0.5 * g * t^2

We know that the initial height, h, is 46 m, and the acceleration due to gravity, g, is approximately 9.8 m/s^2.

Plugging these values into the equation, we get:

46 = 0.5 * 9.8 * t^2

Solving for t, we find:

t^2 = 46 / (0.5 * 9.8)
t^2 = 9.3878
t ≈ 3.06 seconds

So it will take approximately 3.06 seconds for the egg to reach the ground. Now let's calculate how far the professor will walk during that time.

Distance = Speed * Time
Distance = 1.20 * 3.06
Distance ≈ 3.67 meters

Therefore, the professor should be approximately 3.67 meters away from the base of the building when you release the egg. And with a little bit of "egg-curacy" and aim, you'll be able to give your professor a "yolk-y surprise"! Please don't actually drop eggs on your professor, though. It's just a "yolk".

To determine where the professor should be when you release the egg, you need to consider the time it takes for the egg to fall from the roof to the ground.

The equation of motion for a falling object can be described by the kinematic equation:

h = initial_height - (1/2) * g * t^2

Where:
h = height of the object above the ground (in this case, the roof height)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes for the object to fall

Let's solve for t:

h = 46 m
g = 9.8 m/s^2

46 = 0 - (1/2) * 9.8 * t^2

Rearrange the equation to solve for t:

-46 = -4.9 * t^2

Divide both sides by -4.9:

t^2 = 9.3878

Take the square root of both sides to solve for t:

t = √9.3878 ≈ 3.06 s

Now, we have the time it takes for the egg to fall. The distance the professor will have walked during this time can be calculated by multiplying the professor's walking speed by the duration:

distance = speed * time
= 1.20 m/s * 3.06 s
≈ 3.67 m

Therefore, the professor should be approximately 3.67 meters away from the base of the building when you release the egg if you wish to hit their head.