/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A rocket engine can accelerate a... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 69

A rocket engine can accelerate a rocket launched from rest vertically up with an acceleration of \(20.0 \mathrm{m} / \mathrm{s}^{2}\) However, after 50.0 s of flight the engine fails. (a) What is the rocket's altitude when the engine fails? (b) When does it reach its maximum height? (c) What is the maximum height reached? [Hint: A graphical solution may be easiest.] \((\mathrm{d})\) What is the velocity of the rocket just before it hits the ground?

Short Answer

Expert verified
Based on the given information and calculations: (a) The rocket's altitude when the engine fails is 25,000 meters. (b) The time it takes for the rocket to reach maximum height is approximately 102.04 seconds. (c) The maximum height reached by the rocket is approximately 26,510.2 meters. (d) The velocity of the rocket just before hitting the ground is approximately 718.92 m/s downward.

Step by step solution

01

Initial conditions

The problem states that the rocket is initially at rest and accelerates vertically upwards with an acceleration of \(20.0 \, \mathrm{m/s^2}\). So we have initial velocity \(v_0 = 0 \, \mathrm{m/s}\) and acceleration \(a = 20.0 \, \mathrm{m/s^2}\). The engine fails after 50 seconds.
02

Compute the rocket's position and velocity after 50 seconds

First, we will find the final velocity \(v_f\) at the end of 50 seconds, using the formula: \(v_f = v_0 + a \cdot t\), where \(t = 50 \, \mathrm{s}\). In this case, \(v_f = 0 + 20.0 \cdot 50 = 1000 \, \mathrm{m/s}\). Next, we will find the altitude \(y\) at that moment, using the formula: \(y = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\). In this case, \(y = 0 + \frac{1}{2} \cdot 20.0 \cdot (50)^2 = 25000 \, \mathrm{m}\).
03

Compute the maximum height

When the engine fails, the rocket will continue to move upwards due to its inertia, before gravity eventually causes it to fall back down. At the maximum height, the rocket's velocity will be \(0 \, \mathrm{m/s}\). Let's denote the time it takes to reach the maximum height as \(t_m\). We'll use the formula \(v = v_0 + a \cdot t\) again, but this time, \(v = 0, v_0 = 1000 \, \mathrm{m/s}\), and \(a = -9.8 \, \mathrm{m/s^2}\). We need to find \(t_m\) from this equation, and then we can find the maximum height.
04

Find the time to reach the maximum height

Solving for \(t_m\) from the equation: \(0 = 1000 - 9.8 \cdot t_m \Rightarrow t_m = \frac{1000}{9.8} \approx 102.04 \, \mathrm{s}\). Now we can find the maximum height using the formula: \(h_m = y + v_0' \cdot t_m + \frac{1}{2}\cdot a' \cdot (t_m)^2\). Note that \(v_0' = 1000 \, \mathrm{m/s}\) and \(a' = -9.8 \, \mathrm{m/s^2}\).
05

Calculate the maximum height

Plugging in the values into the formula: \(h_m = 25000 + 1000 \cdot 102.04 - \frac{1}{2} \cdot 9.8 \cdot (102.04)^2 \approx 26510.2 \, \mathrm{m}\). So the maximum height is approximately 26510.2 meters.
06

Calculate the time it takes for the rocket to fall back to the ground

Let's denote the time it takes for the rocket to fall back to the ground as \(t_f\). Since the rocket's final position is back on the ground \((y = 0)\), we can use the formula: \(y = v_0 \cdot t_f + \frac{1}{2} \cdot a \cdot (t_f)^2\), with \(y = 0, v_0 = 0 \, \mathrm{m/s}\), and \(a = -9.8 \, \mathrm{m/s^2}\). We need to find \(t_f\) from this equation and then find the velocity just before hitting the ground.
07

Calculate the total time for the rocket to hit the ground

Solve for \(t_f\) in the formula: \(0 = -26510.2 + 4.9 \cdot (t_f)^2 \Rightarrow t_f = \sqrt{\frac{26510.2}{4.9}} \approx 73.36 \, \mathrm{s}\). Now, we can find the velocity just before hitting the ground using the formula: \(v_f = v_0 + a \cdot t_f\). Remember that \(v_0 = 0 \, \mathrm{m/s}\) and \(a = -9.8 \, \mathrm{m/s^2}\).
08

Calculate the velocity before hitting the ground

Plugging in the values into the formula: \(v_f = 0 -9.8 \cdot 73.36 = -718.92 \, \mathrm{m/s}\). So the velocity just before the rocket hits the ground is approximately \(718.92 \, \mathrm{m/s}\) downward. #Answers# (a) The rocket's altitude when the engine fails is 25000 meters. (b) The time to reach maximum height is approximately 102.04 seconds. (c) The maximum height reached is approximately 26510.2 meters. (d) The velocity just before the rocket hits the ground is approximately 718.92 m/s downward.

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