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Chapter 2: Problem 54

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A student, looking toward his fourth-floor dormitory window, sees a flowerpot with nasturtiums (originally on a window sill above) pass his \(2.0-\mathrm{m}\) high window in 0.093 s. The distance between floors in the dormitory is $4.0 \mathrm{m} .$ From a window on which floor did the flowerpot fall?

Short Answer

Expert verified
Answer: The flowerpot was dropped from the fourth-floor window.

Step by step solution

01

Identify given data

Here are the given data: - Time taken to pass the window: \(t = 0.093 \ \text{s}\) - Height of the student's window: \(h_\text{window} = 2.0 \ \text{m}\) - Distance between floors: \(d_\text{floors} = 4.0 \ \text{m}\) - Free-fall acceleration: \(g = 9.80 \ \frac{\text{m}}{\text{s}^2}\)
02

Calculate the velocity when the flowerpot reaches the top of the window

We'll use the following kinematic equation to find the velocity \(v_\text{top}\) when the flowerpot reaches the top of the window: \(v^2 = u^2 + 2as\) Where: - \(v\) is the final velocity - \(u\) is the initial velocity (in this case, 0 m/s as it starts from rest) - \(a\) is the acceleration (in this case, -g, as the flowerpot is in free-fall) - \(s\) is the displacement (in this case, the height of the window) \(v_\text{top}^2 = 0^2 + 2(-g)(h_\text{window})\) \(v_\text{top} = \sqrt{2(-g)(h_\text{window})}\) \(v_\text{top} = \sqrt{2(-(-9.80))(2.0)}\) \(v_\text{top} = \sqrt{39.2}\) \(v_\text{top} \approx 6.26 \ \frac{\text{m}}{\text{s}}\)
03

Calculate the total height the flowerpot fell from

Now we'll use another kinematic equation to calculate the height \(h_\text{total}\), taking the calculated velocity at the top of the window: \(h_\text{total} = ut + \frac{1}{2}at^2\) Where: - \(h_\text{total}\) is the total height fallen - \(u\) is the initial velocity (in this case, 0 m/s as it starts from rest) - \(t\) is the time taken to pass the window - \(a\) is the acceleration (in this case, -g, as the flowerpot is in free-fall) \(h_\text{total} = 0 \cdot t + \frac{1}{2}(-g)t^2\) \(h_\text{total} = -\frac{1}{2}(9.80)(0.093)^2\) \(h_\text{total} \approx 0.0423 \ \text{m}\)
04

Determine the initial height where it started falling from

Now, adding the height of the window to the height calculated in step 3, we'll find the initial height: \(h_\text{initial} = h_\text{total} + h_\text{window}\) \(h_\text{initial} = 0.0423 + 2.0\) \(h_\text{initial} \approx 2.0423 \ \text{m}\)
05

Determine the floor the flowerpot fell from

Now, we'll divide the initial height by the distance between the floors: \(N_\text{floors} = \frac{h_\text{initial}}{d_\text{floors}}\) \(N_\text{floors} = \frac{2.0423}{4.0}\) \(N_\text{floors} \approx 0.5106\) Since the result is about half a floor above the student's floor (the 0.5106 value), the flowerpot must have fallen from one floor above the student, so it was dropped from his fourth-floor window.

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Most popular questions from this chapter

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0 -m cliff. It leaves his hands with zero speed, and after \(2.0 \mathrm{s}\) it has attained a velocity of \(3.3 \mathrm{m} / \mathrm{s}\) downward. How far has the camera fallen after \(4.0 \mathrm{s} ?\)

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You are driving your car along a country road at a speed of $27.0 \mathrm{m} / \mathrm{s} .$ As you come over the crest of a hill, you notice a farm tractor \(25.0 \mathrm{m}\) ahead of you on the road, moving in the same direction as you at a speed of \(10.0 \mathrm{m} / \mathrm{s} .\) You immediately slam on your brakes and slow down with a constant acceleration of magnitude $7.00 \mathrm{m} / \mathrm{s}^{2} .$ Will you hit the tractor before you stop? How far will you travel before you stop or collide with the tractor? If you stop, how far is the tractor in front of you when you finally stop?
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