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

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The discussion focuses on calculating the precise position of a teacher walking towards a building to ensure an egg dropped from a height of 46.0 meters lands on their head. The teacher, who is 1.80 meters tall, must be positioned 3.6 meters away from the building when the egg is released. The calculations involve determining the fall time of the egg, which is 3 seconds, and the distance the teacher covers while walking at a speed of 1.20 m/s. The key equations used include kinematic equations for vertical and horizontal motion.

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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 [itex]x = v_{y}_{0}t + \frac{1}{2} a_{x}t^{2}[/itex] for the egg, and [itex]x = x_{0} + v_{x}_{0}t + \frac{1}{2} a_{x}t^{2}[/itex]?

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

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