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Chapter 2: Problem 62

In the problems, 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 stone is thrown vertically downward from the roof of a building. It passes a window \(16.0 \mathrm{m}\) below the roof with a speed of $25.0 \mathrm{m} / \mathrm{s} .\( It lands on the ground \)3.00 \mathrm{s}$ after it was thrown. What was (a) the initial velocity of the stone and (b) how tall is the building?

Short Answer

Expert verified
Answer: The initial velocity of the stone is 33.0 m/s (downward), and the height of the building is 59.1 m.

Step by step solution

01

Define the variables

Let's define the following variables: - \(x_{0}\) 鈥� distance from the roof to the window (16 meters) - \(v_{f}\) 鈥� final velocity of the stone when it passes the window (25 m/s) - \(t\) 鈥� time when the stone passes the window - \(v_{0}\) 鈥� initial velocity of the stone (unknown) - \(g = 9.8\) \(m/s^2\) 鈥� acceleration due to gravity (negative, as it is directed downward) - \(h\) 鈥� total height of the building (unknown)
02

Find the time when the stone passes the window

We will use the following kinematic equation to find the time when the stone passes the window, sparing \(v_{0}\) for now: \(v_{f} = v_{0} - gt\) Rearrange to isolate \(t\): \(t = \frac{v_{0} - v_{f}}{g}\)
03

Use the displacement equation to find initial velocity

Now use the kinematic equation relating displacement, initial velocity, time, and acceleration: \(x_{0} = v_{0}t - \frac{1}{2}gt^2\) Substitute the value of \(t\) from Step 2 into the equation: \(x_{0} = v_{0}(\frac{v_{0} - v_{f}}{g}) - \frac{1}{2}g(\frac{v_{0} - v_{f}}{g})^2\) Solve for \(v_{0}\) from the above equation, we get: \(v_{0} = 33.0\, \mathrm{m/s}\)
04

Find the height of the building

Now that we know the initial velocity, we will use the time, given as \(3\, \mathrm{s}\) to find the height of the building. Use the displacement equation again: \(h = v_{0}t - \frac{1}{2}gt^2\) Plug in the values (\(v_{0} = 33.0\, \mathrm{m/s}\) , \(t = 3\, \mathrm{s}\) , and \(g = 9.8\, \mathrm{m/s^2}\)): \(h = (33.0)(3) - \frac{1}{2}(9.8)(3)^2\) Solve for \(h\): \(h = 59.1\, \mathrm{m}\)
05

Final Answers

(a) The initial velocity of the stone is \(33.0\, \mathrm{m/s}\) (downward). (b) The height of the building is \(59.1\, \mathrm{m}\).

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