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Chapter 2: Problem 45

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 penny is dropped from the observation deck of the Empire State building (369 m above ground). With what velocity does it strike the ground?

Short Answer

Expert verified
Answer: The penny strikes the ground with a velocity of -85.0 m/s (downward).

Step by step solution

01

Identifying the known values

We are given: 1. Initial height, \(h=369\) m 2. Initial velocity, \(v_0=0\) m/s (the penny is dropped from rest) 3. Acceleration due to gravity, \(g = -9.80\,\mathrm{m/s^2}\) (downward acceleration is negative) We want to find the final velocity, \(v\), of the penny when it reaches the ground.
02

Selecting the appropriate kinematic equation

There are several kinematic equations that we can choose from, but in this case, the most appropriate one is: \(v^2 = v_0^2 + 2as\) where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(s\) is the displacement.
03

Plugging in the known values

Since the penny falls downward, the displacement is negative, so \(s = -h = -369\,\mathrm{m}\). \(v^2 = 0^2 + 2\times(-9.80\,\mathrm{m/s^2})\times(-369\,\mathrm{m})\)
04

Solving for the final velocity

Solve for \(v\) by performing the operations: \(v^2 = 2\times 9.80\,\mathrm{m/s^2} \times 369\,\mathrm{m}\) \(v^2 = 7225.6\) Now, to find the final velocity, take the square root of both sides: \(v = \sqrt{7225.6}= \pm 85.0\,\mathrm{m/s}\) Since the penny is moving downward when it hits the ground, we take the negative value: \(v = -85.0\,\mathrm{m/s}\)
05

Write down the final answer

The penny strikes the ground with a velocity of \(-85.0\,\mathrm{m/s}\) (downward).

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