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Chapter 2: Problem 61

A dog sees a flowerpot sail up and then back down past a window \(1.1 \mathrm{~m}\) high. If the total time the pot is in sight is \(0.54 \mathrm{~s}\), find the height above the top of the window to which the pot rises.

Short Answer

Expert verified
The height above the top of the window to which the pot rises can be found by following the steps outlined above: Identifying the given quantities, using the 2nd equation of motion to find total distance covered, and finding the height by subtracting the window's height from half the total distance covered.

Step by step solution

01

Identify Knowns and Unknowns

First, values that are known or given by the problem are identified. In this problem, we know the time (\(t\)) which is 0.54 seconds, the acceleration due to gravity (\(a\)) which is \(-9.8 m/s^2\), and the height of the window (\(h\)) which is 1.1 meters. The unknown quantity to find is the maximum height above the window that the pot reaches.
02

Use the 2nd Equation of Motion

Next, the 2nd equation of motion is used: \(s = ut + \frac{1}{2} a t^2\), where \(u\) is the initial velocity of the pot, \(a\) is the acceleration, and \(t\) is the time. Given that at its maximum height, the pot's velocity (\(u\)) is zero, the equation simplifies to: \(s = \frac{1}{2} a t^2\). The total distance \(s\) covered by the pot can then be calculated as follows: \(s = \frac{1}{2} * -9.8 m/s^2 * (0.54 s)^2\).
03

Find the height above the Window

It's important to remember that the distance \(s\) calculated in the previous step is the total distance the pot travelled when in sight. Since the pot passed the window twice (going up and then coming down), the distance above the window should be half of the total distance covered. Therefore, the height above the window is calculated by subtracting the height of the window from half the total distance covered: \(Height = \frac{s}{2} - h\).

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Most popular questions from this chapter

(a) With what speed must a ball be thrown vertically up in order to rise to a maximum height of \(53.7 \mathrm{~m} ?(b)\) For how long will it be in the air?

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