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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 are trying to improve your shooting skills by shooting at a can on top of a fence post. You miss the can, and the bullet, moving at \(200 . \mathrm{m} / \mathrm{s},\) is embedded \(1.5 \mathrm{~cm}\) into the post when it comes to a stop. If constant acceleration is assumed, how long does it take for the bullet to stop?

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