You attach a meter stick to an oak tree, such that the top of the meter stick is 2.07 meters above the ground. Later, an acornfalls from somewhere higher up in the tree. If the acorn takes 0.306 seconds to pass the length of the meter stick, how highho above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on theway down?ho* TOOLSx10

Answered: You attach a meter stick to an oak…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.07 meters above the ground. Later, an acorn
falls from somewhere higher up in the tree. If the acorn takes 0.306 seconds to pass the length of the meter stick, how high
ho above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the
way down?
ho
* TOOLS
x10

Expert Solution

Step 1 :Introduction

The diagram of the system is given below:

According to the Newton's law of motion , the vertical distance covered by an object initially at rest is calculated by using the formula $h = \frac{1}{2} g t^{2}$ , where $g = 9.8 m / s^{2}$ is the acceleration due to gravity and t is the time taken.

When the acorn reaches the top of the scale , the formula will be $h_{0} - 2.07 m = \frac{1}{2} g {t_{1}}^{2} \to (1)$ , where $h_{0}$ is the initial height of the acorn and $t_{1}$ is the time taken to reach the height $h = 2.07 m$ .
When the acorn reaches the bottom of the scale , the formula will be $h_{0} - 1.07 m = \frac{1}{2} g {t_{2}}^{2} \to (2)$ , where $h_{0}$ is the initial height of the acorn and $t_{2}$ is the time taken to reach the height $h' = 1.07 m$ , where the time taken to cross the meter scale is given as $t_{2} - t_{1} = 0.306 s$ .

Substract the equation (1) from (2) , thus ,

$\begin{array}{rcl} (h_{0} - 2.07 m) - (h_{0} - 1.07 m) & = & \frac{1}{2} g ({t_{1}}^{2} - {t_{2}}^{2}) \\ - 1 m & = & \frac{1}{2} \times (9.8 m / s^{2}) \times ({t_{1}}^{2} - {t_{2}}^{2}) \\ 1 m & = & (4.9 m / s^{2}) \times ({t_{2}}^{2} - {t_{1}}^{2}) \\ (t_{2} - t_{1}) (t_{2} + t_{1}) & \approx & 0.204 s^{2} \\ t_{2} + t_{1} & \approx & \frac{0.204 s^{2}}{0.306 s} \\ \approx & 0.7 s \end{array}$

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