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Chapter 3: Problem 54

What is the maximum height above ground a projectile of mass \(0.79 \mathrm{~kg}\), launched from ground level, can achieve if you are able to give it an initial speed of \(80.3 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The maximum height the projectile can achieve is approximately 261.52 meters above ground level.

Step by step solution

01

Write down the conservation of energy equation

Since the energy is conserved, the initial kinetic energy (KE) will equal the potential energy (PE) at the maximum height. We can write this as: \(KE_{initial} = PE_{max}\)
02

Write the expressions for initial kinetic energy and potential energy at the maximum height

The initial kinetic energy can be expressed as: \(KE_{initial} = \frac{1}{2}mv^2\) And potential energy at the maximum height can be expressed as: \(PE_{max} = mgh_{max}\) Here, \(m\) is the mass of the projectile, \(v\) is the initial speed, \(g\) is the acceleration due to gravity, and \(h_{max}\) is the maximum height.
03

Substitute the expressions and given values into the conservation of energy equation

Recall the conservation of energy equation we wrote in Step 1: \(KE_{initial} = PE_{max}\) Now, substitute the expressions from Step 2 and the given values of mass and initial speed: \(\frac{1}{2}(0.79\mathrm{kg})(80.3 \mathrm{m/s})^2 = (0.79\mathrm{kg})(9.81 \mathrm{m/s^2})h_{max}\)
04

Solve for the maximum height \(h_{max}\)

Now we only have one unknown in the equation, which is \(h_{max}\). Solve for \(h_{max}\): \(h_{max} = \frac{\frac{1}{2}(0.79)(80.3)^2}{(0.79)(9.81)}\) \(h_{max} \approx 261.52 \mathrm{m}\) Thus, the maximum height the projectile can achieve is approximately 261.52 meters above ground level.

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