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Chapter 12: Problem 7

A 0.2 -lb model rocket is launched vertically from rest at time \(t=0\) with a constant thrust of 2 lb for one second and no thrust for \(t>1\) s. Neglecting air resistance and the decrease in mass of the rocket, determine ( \(a\) ) the maximum height \(h\) reached by the rocket, \((b)\) the time required to reach this maximum height.

Short Answer

Expert verified
(a) Maximum height is 1453.63 ft. (b) Time to reach maximum height is 10.02 s.

Step by step solution

01

Convert weight to mass

The model rocket's weight is given as 0.2 lb. To find the mass, we use the relation between weight \( W \) and mass \( m \), given by \( W = mg \), where \( g = 32.2 \, \text{ft/s}^2 \) (standard gravity). Thus, the mass \( m = \frac{W}{g} = \frac{0.2}{32.2} \approx 0.0062 \, \text{slug}.\)
02

Calculate Rocket's Initial Acceleration

The thrust provided by the rocket is 2 lb. Using Newton's second law \( F = ma \), where \( F \) is the net force and \( a \) the acceleration, calculate the net force: \( F_{net} = 2 - 0.2 = 1.8 \) lb (since 0.2 lb is the weight). Thus, the net acceleration \( a = \frac{F_{net}}{m} = \frac{1.8}{0.0062} \approx 290.32 \, \text{ft/s}^2.\)
03

Determine Velocity after Thrust Duration

The thrust is applied for 1 second. Using the equation of motion \( v = u + at \), where initial velocity \( u = 0 \), \( a = 290.32 \, \text{ft/s}^2 \), and \( t = 1 \) s, the final velocity \( v = 0 + 290.32 \times 1 \approx 290.32 \, \text{ft/s}.\)
04

Calculate Maximum Height

For the upward motion after thrust ends, use the equation \( v^2 = u^2 + 2as \) with \( v = 0 \), \( u = 290.32 \, \text{ft/s} \), and \( a = -32.2 \, \text{ft/s}^2 \) (gravity slowing it down) to solve for \(s\), the additional height: \( 0 = 290.32^2 + 2(-32.2)s \). Simplifying, \( s = \frac{290.32^2}{2 \times 32.2} \approx 1308.47 \, \text{ft}.\) Adding the height obtained during thrust (using \( s = ut + 0.5at^2 = 0 + 0.5 \times 290.32 \times 1^2 \)) which equals 145.16 ft, the total height \( h = 1308.47 + 145.16 \approx 1453.63 \, \text{ft}.\)
05

Calculate Time to Reach Maximum Height

Using \( v = u + at \) after thrust ends, where \( v = 0 \) at max height, \( u = 290.32 \, \text{ft/s} \), and \( a = -32.2 \, \text{ft/s}^2 \), solve for \( t \): \( 0 = 290.32 - 32.2t \). Thus, \( t = \frac{290.32}{32.2} \approx 9.02 \, \text{s}.\) Including the thrust phase, the total time to peak is \( 1 + 9.02 = 10.02 \, \text{s}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law tells us how an object will move when acted upon by a force. The law is often stated as:\[ F = ma \]where \( F \) represents force, \( m \) is the mass of the object, and \( a \) is the acceleration. In this exercise, we used the second law to find the net acceleration of a rocket.
  • The rocket's thrust was 2 lbs, and its weight was 0.2 lbs, providing a net force of 1.8 lbs upward.
  • This net force causes the rocket to accelerate upwards at \( \frac{1.8}{0.0062} \approx 290.32 \, \text{ft/s}^2 \).
Newton's Second Law allowed us to calculate how quickly the rocket speeds up, which is key to solving for the flight's further characteristics.
Kinematics Equations
Kinematics equations help us understand the motion of objects. These equations relate the object's velocity, acceleration, and displacement over time. For our rocket's motion, we used the equation:\[ v = u + at \]where \( v \) is the final velocity, \( u \) the initial velocity, \( a \) the acceleration, and \( t \) the time. During the thrust phase, this equation helped calculate the velocity reached:
  • The initial velocity \( u \) was 0, and the acceleration \( a \) was calculated at \( 290.32 \, \text{ft/s}^2 \).
  • With the thrust lasting 1 second, the rocket's velocity became \( v = 290.32 \, \text{ft/s} \).
By applying kinematics, we predicted the motion throughout the rocket's flight phase, including the slowing period after thrust.
Maximum Height Calculation
The maximum height calculation uses kinematics to determine how high the rocket flies after thrust ends. Here, the equation for motion under constant acceleration came into play:\[ v^2 = u^2 + 2as \]In this scenario, once the thrust ends, gravity acts as a decelerating force with \( a = -32.2 \, \text{ft/s}^2 \). We used:
  • \( u = 290.32 \, \text{ft/s} \) as the velocity at the end of the thrust.
  • Setting \( v = 0 \) to find when the rocket stops rising.
  • The calculation revealed a height \( s \approx 1308.47 \, \text{ft} \) reached after thrust.
Adding this to the rise during thrust gave us a total maximum height of approximately 1453.63 ft.
Thrust and Acceleration
Thrust and acceleration are central to understanding how the rocket moves. Thrust refers to the force propelling the rocket upwards. An object's response to this force is acceleration. Here's how it worked in this problem:
  • The thrust of 2 lbs propelled the rocket upward.
  • Subsequent acceleration was determined by dividing the net force (1.8 lbs) by the rocket's mass (0.0062 slug).
The thrust lasted for 1 second, resulting in rapid acceleration and an upward velocity reaching 290.32 ft/s. Once the thrust ended, the rocket continued upward at this velocity until gravity gradually slowed it down. Understanding these elements is vital to figuring out the rocket's trajectory and final height.

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