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Chapter 2: Problem 71

An object is thrown vertically and has an upward velocity of \(25 \mathrm{~m} / \mathrm{s}\) when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

Short Answer

Expert verified
Answer: The initial speed of the object is approximately 35.36 m/s.

Step by step solution

01

Write down the given information

We are given that - The object's upward velocity is 25 m/s when it reaches 1/4 of its maximum height - We need to find the initial (launch) speed of the object
02

Apply the Kinematic Equations

To solve this problem, we can use the following kinematic equation for vertical motion: $$ v^2 = u^2 + 2as $$ where - \(v\) is the final velocity, which is given at 25 m/s - \(u\) is the initial velocity that we need to find - \(a\) is the acceleration, which is the acceleration due to gravity (\(-9.81 \mathrm{~m/s^2}\)) - \(s\) is the displacement(vertical distance covered), which is given as 1/4 of the object's maximum height
03

Rewrite the displacement

The maximum height of an object thrown vertically is reached when its final velocity, \(v_f\), becomes 0. We can use the kinematic equation again: $$ v_f^2 = u^2 + 2as $$ to find the maximum height, \(s_{max}\), using the initial velocity \(u\) that we are looking for, the acceleration due to gravity, \(a = -9.81 \mathrm{~m/s^2}\), and displacement, \(s_{max}\). As \(v_f=0\), we have $$ 0 = u^2 - 2 \times 9.81 \times s_{max} $$ This means $$ s_{max} = \frac{u^2}{2 \times 9.81} $$ We are also given that the object has a velocity of \(25 \mathrm{~m/s}\) when it reaches 1/4 of its maximum height, \(s = \frac{1}{4}s_{max}\).
04

Write down equation for given velocity

Now we can write the equation for the given velocity \(v=25 \mathrm{~m/s}\): $$ (25 \mathrm{~m/s})^2 = u^2 - 2 \times 9.81 \times \frac{1}{4}s_{max} $$ Substitute the expression for \(s_{max}\) as found in Step 3: $$ (25 \mathrm{~m/s})^2 = u^2 - 2 \times 9.81 \times \frac{1}{4} \times \frac{u^2}{2 \times 9.81} $$
05

Solve for the initial velocity

Simplify the equation and solve for \(u\): $$ (25 \mathrm{~m/s})^2 = u^2 - \frac{1}{2}u^2 $$ $$ (25 \mathrm{~m/s})^2 = \frac{1}{2}u^2 $$ $$ u^2=\frac{2\times(25 \mathrm{~m/s})^2}{1} $$ Therefore, $$ u= \sqrt{\frac{2\times(25 \mathrm{~m/s})^2}{1}} = 35.36 \mathrm{~m/s} $$ So, the initial (launch) speed of the object is approximately \(35.36 \mathrm{~m/s}\).

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Most popular questions from this chapter

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