Placing x-markers to drop water balloon on students

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SUMMARY

The discussion focuses on calculating the distance to place an x-marker for dropping water balloons from a 20m high window onto students approaching at a speed of 2 m/s. The correct free-fall time was computed, leading to a distance of approximately 4.03m from the impact point. The confusion arose from mislabeling units, mistaking velocity for acceleration, which was clarified in the conversation. The final calculation utilized the formula for distance with constant velocity and zero acceleration.

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Homework Statement



You and your roommate plot to drop water balloons on student entering your dorm. your window is 20m above the side walk. You plan is to place an x on the side walk to mark the spot a student must be when you drop the balloon. You note that most student approach the dorm at 2ms^-1. How far from the impact point do you place the x?

Homework Equations



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The Attempt at a Solution



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The first half, where you computed the free falling time, is correct. The second half - where does 2 ms-2 come from?
 
voko said:
The first half, where you computed the free falling time, is correct. The second half - where does 2 ms-2 come from?

Opps. A slip in the label of unit which had me confuse velocity for acceleration. I'll redo it. Can I assume acceleration is 0ms^-2? I think I could since the walking velocity is constant, acceleration is zero.
 
xf - xi = 2ms^-1 (2.0192751098s) + 0.5(0ms^-2)(2.019275109s)^2 = 4.03855m
 
Good, but you have way too many "significant" figures.
 
voko said:
Good, but you have way too many "significant" figures.

rounded off to 4.03m
 

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