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Chapter 17: Problem 16

A projectile is shot straight up with \(v_{o}=40 \mathrm{~m} / \mathrm{s}\). After how many seconds will it return if drag is neglected? a) \(4 \mathrm{~s}\) c) \(8 \mathrm{~s}\) b) \(6 \mathrm{~s}\) d) \(10 \mathrm{~s}\)

Short Answer

Expert verified
Question: A projectile is shot straight up from the ground with an initial velocity of 40 m/s. Neglecting drag, choose the approximate time (in seconds) it takes for the projectile to return to its starting point. a) 5 s b) 7 s c) 8 s d) 10 s Answer: c) 8 s

Step by step solution

01

Identify the given information

We are given the initial velocity \(v_{o} = 40 \mathrm{~m} / \mathrm{s}\), and we need to find the time it takes for the projectile to return to its starting point, neglecting drag.
02

Use the equation of motion to find the time when the projectile reaches maximum height

At the maximum height, the velocity will be 0. We can use the equation \(v = v_{o} - gt\) to find the time it takes to reach maximum height, where \(v\) is the final velocity, \(v_{o}\) is the initial velocity, \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m} / \mathrm{s}^2\)), and \(t\) is the time. We can rearrange the equation to solve for \(t\): \(t = \frac{v_{o} - v}{g}\) Plug in the given values: \(t = \frac{40 \mathrm{~m} / \mathrm{s} - 0 \mathrm{~m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^2}\)
03

Calculate the time when the projectile reaches maximum height

Now, compute the time it takes to reach the maximum height: \(t = \frac{40 \mathrm{~m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^2} = 4.08 \mathrm{~s}\).
04

Calculate the total time for the projectile to return

Since the path is symmetrical, the total time it takes for the projectile to return is twice the time it takes to reach the maximum height: Total time = \(2 \times 4.08 \mathrm{~s} = 8.16 \mathrm{~s}\).
05

Choose the closest answer from the options provided

From the available options, the closest approximate answer to the calculated time is: c) \(8 \mathrm{~s}\)

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