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Chapter 2: Problem 48

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 \(4.0 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: Approximately 13.2 meters.

Step by step solution

01

Identify the relevant equations

We will use the following two equations for this exercise: 1. Displacement: \(s = ut + \frac{1}{2}at^2\), where \(s\) is the displacement, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration. 2. Velocity: \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration.
02

Calculate the acceleration on the Moon

We have the velocity after 2.0 seconds, and we know that the initial velocity is 0.0 m/s. We can use the velocity equation to calculate the acceleration. \(v = u + at\) \(3.3 \mathrm{m} / \mathrm{s} = 0.0 \mathrm{m} / \mathrm{s} + a(2.0 \mathrm{s})\) Now we can solve for the acceleration, a: \(a = \frac{3.3 \mathrm{m} / \mathrm{s}}{2.0 \mathrm{s}} = 1.65 \mathrm{m} / \mathrm{s}^{2}\)
03

Determine the displacement after 4.0 seconds

Now that we have the acceleration, we can use the displacement equation to find how far the camera has fallen after 4.0 seconds. \(s = ut + \frac{1}{2}at^2\) Since the initial velocity, \(u\), is 0.0 m/s, the equation becomes: \(s = \frac{1}{2}at^2\) Now, we can plug in the values for the acceleration and time: \(s = \frac{1}{2}(1.65 \mathrm{m} / \mathrm{s}^{2})(4.0 \mathrm{s})^2\) \(s \approx 13.2 \mathrm{m}\) So after 4.0 seconds, the camera has fallen approximately 13.2 meters.

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