Where should the teacher be for the egg to drop on their head?

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To determine where the teacher should be for the egg to drop on his head, the height from which the egg falls is calculated as 44.2 meters, accounting for the teacher's height of 1.8 meters. The time it takes for the egg to fall this distance is approximately 3 seconds, derived from the equation of motion. During this time, the teacher, moving at a speed of 1.20 m/s, will cover a distance of 3.6 meters. Therefore, the teacher needs to be positioned 3.6 meters away from the building when the egg is released. This ensures the egg lands directly on the teacher's head.
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A person is on the roof of a building, 46.0 m above the ground. His teacher, who is 1.80 meters tall, is walking toward the building at a constant speed of 1.20 m/s. If the person wishes to drop the egg on his teacher's head, where should the teacher be when the person releases the egg?

So I know that I need to set two equations equal to each other to get the position the teacher needs to be for the egg to drop on him. So should I use x = v_{y}_{0}t + \frac{1}{2} a_{x}t^{2} for the egg, and x = x_{0} + v_{x}_{0}t + \frac{1}{2} a_{x}t^{2}?

Thanks
 
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How far does the egg have to fall before impacting on the head of the 1.8 m tall teacher ?
How long will it take for the egg to fall this distance ?
How far can teacher walk in this time ?
 
Thanks. I got the answer.

1. 46 - 1.8 = 44.2 m
2. 44.2 = \frac{1}{2} (9.8) t^{2}, t = 3
3. x = 0 + 1.20(3) = 3.6 m
 
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