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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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