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Chapter 2: Problem 79

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). What maximum height above the projection point does it reach?

Short Answer

Expert verified
Answer: The maximum height reached by the object is 40 m.

Step by step solution

01

Identify the known variables and the unknown variable.

We know the initial upward velocity (\(v_0\)) is \(28.0 \mathrm{~m} / \mathrm{s}\). We want to find the maximum height reached (\(h\)). We also know that when it reaches the maximum height, its final velocity (\(v\)) will be \(0 \mathrm{~m} / \mathrm{s}\). The acceleration (\(a\)) is due to gravity, so \(a = -9.8 \mathrm{~m} / \mathrm{s}^2\).
02

Choose the appropriate equation of motion.

Since we know the initial and final velocities, acceleration, and we want to find the maximum height, we can use the second equation of motion: \(v^2 = v_0^2 + 2 a h\).
03

Substitute the known values into the chosen equation and rearrange for the unknown variable.

Let's substitute the known values of \(v_0\), \(v\), and \(a\) into the equation and solve for \(h\). The equation becomes: \(0^2 = (28.0 \mathrm{~m} / \mathrm{s})^2 + 2(-9.8 \mathrm{~m} / \mathrm{s}^2) h\). Rearranging for \(h\), we get \(h = \frac{(28.0 \mathrm{~m} / \mathrm{s})^2 - 0^2}{2(-9.8 \mathrm{~m} / \mathrm{s}^2)}\).
04

Calculate the maximum height reached.

We can now calculate the value of \(h\): \(h = \frac{(28.0 \mathrm{~m} / \mathrm{s})^2}{2(9.8 \mathrm{~m} / \mathrm{s}^2)} = \frac{784 \mathrm{~m}^2 / \mathrm{s}^2}{19.6 \mathrm{~m} / \mathrm{s}^2} = 40 \mathrm{~m}\). Therefore, the maximum height reached by the object above the projection point is \(40 \mathrm{~m}\).

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Most popular questions from this chapter

You drop a rock over the edge of a cliff from a height \(h\). Your friend throws a rock over the edge from the same height with a speed \(v_{0}\) vertically downward, at some time \(t\) after you drop your rock. Both rocks hit the ground at the same time. How long after you dropped your rock did your friend throw hers? Express your answer in terms of \(v_{0}, g,\) and \(h\).

The position of a particle as a function of time is given as \(x(t)=\frac{1}{4} x_{0} e^{3 \alpha t}\), where \(\alpha\) is a positive constant. a) At what time is the particle at \(2 x_{0}\) ? b) What is the speed of the particle as a function of time? c) What is the acceleration of the particle as a function of time? d) What are the SI units for \(\alpha\) ?

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