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Chapter 2: Problem 46

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) How long does it take for a golf ball to fall from rest for a distance of \(12.0 \mathrm{m} ?\) (b) How far would the ball fall in twice that time?

Short Answer

Expert verified
(b) How far does the golf ball fall in twice the time it takes to fall 12 meters? Answer: (a) It takes approximately 1.56 seconds for the golf ball to fall 12 meters. (b) The golf ball falls approximately 47.97 meters in twice the time (3.12 seconds).

Step by step solution

01

Identify the relevant equation of motion

Since the ball is falling freely, we can use the following equation of motion: \(y = v_{0}t + \frac{1}{2}at^2\) where y is the vertical distance, \(v_0\) is the initial velocity, t is the time, a is the vertical acceleration (\(g = 9.80 m/s^2\)), and \(y = 12m\).
02

Solve for time t

Since the ball is falling from rest, the initial velocity \(v_0\) is 0. So the equation of motion becomes: \(12 = 0t + \frac{1}{2}(9.80)t^2\) Solve for t: \(t^2 = \frac{12 * 2}{9.80}\) \(t = \sqrt{\frac{24}{9.80}}\)
03

Calculate the time

Plug in the values and solve for t: \(t = \sqrt{\frac{24}{9.80}} \approx 1.56s\) So it takes approximately 1.56 seconds for the golf ball to fall 12 meters. #b) Distance fallen in twice the time#
04

Calculate twice the time

We found in part (a) that it takes 1.56 seconds for the golf ball to fall 12 meters. Now, we will calculate twice this time: \(2t = 2(1.56) = 3.12s\)
05

Use the equation of motion to find the distance fallen

We will again use the equation of motion to find the distance fallen in 3.12 seconds: \(y = v_{0}t + \frac{1}{2}at^2\) Since the initial velocity \(v_0\) is 0, the equation simplifies to: \(y = \frac{1}{2}(9.80)(3.12)^2\)
06

Calculate the distance fallen

Plug in the values and solve for y: \(y = \frac{1}{2}(9.80)(3.12)^2 \approx 47.97m\) So the golf ball falls approximately 47.97 meters in twice the time (3.12 seconds).

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Most popular questions from this chapter

For the train of Example \(2.2,\) find the average velocity between 3: 14 P.M. when the train is at \(3 \mathrm{km}\) east of the origin and 3: 28 P.M. when it is \(10 \mathrm{km}\) east of the origin.
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