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Chapter 11: Problem 36

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was \(89.6 \mathrm{ft}\) the end of the powered portion of the flight and that the rocket landed \(16 \mathrm{s}\) later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that \(g=32.2 \mathrm{ft} / \mathrm{s}^{2},\) determine \((a)\) the speed \(v_{1}\) of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket.

Short Answer

Expert verified
(a) The speed \(v_1\) is 257.6 ft/s; (b) The maximum altitude is 1120 ft.

Step by step solution

01

Understanding the Problem

We are given the altitude at the end of the powered flight (89.6 ft) and the total time of free fall (16 seconds) to solve for the rocket's speed at the end of the powered flight and its maximum altitude. The force of gravity acting on the rocket during free fall is provided as \(g = 32.2 \mathrm{ft/s^2}\).
02

Calculate Time to Reach Maximum Altitude

During free fall, the rocket reaches its maximum altitude before falling downward. The time to reach maximum altitude (\(t_{up}\)) can be found by dividing the total free-fall time by 2, since the ascent and descent times are equal: \[ t_{up} = \frac{16}{2} = 8 \text{ seconds} \]
03

Determine Initial Speed at End of Powered Flight

Using the kinematic equation for the velocity at the maximum point, which is zero, and knowing the time to climb \(t_{up}\) and acceleration due to gravity \(g\): \[ v_{1} = g \cdot t_{up} = 32.2 \cdot 8 = 257.6 \text{ ft/s} \]This is the speed of the rocket at the end of powered flight.
04

Calculate Maximum Altitude Reached

Use the kinematic equation to find the additional altitude gained during the upward motion until maximum height is reached: \[ h_{additional} = v_{1} \cdot t_{up} - \frac{1}{2} \cdot g \cdot t_{up}^2 \]\[ h_{additional} = 257.6 \cdot 8 - \frac{1}{2} \cdot 32.2 \cdot 8^2 \]\[ h_{additional} = 2060.8 - 1030.4 = 1030.4 \text{ ft} \]Add this to the altitude at the end of powered flight to get the max altitude: \[ h_{max} = 89.6 + 1030.4 = 1120 \text{ ft} \]
05

Verify Calculations

Review each step to ensure calculations and formulas were correctly applied. Ensure units match physical quantities' requirements.

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Key Concepts

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

Kinematic Equations
Kinematic equations are key tools in understanding motion. When studying rocket motion, these equations allow us to calculate various motion aspects such as velocity, displacement, and time. The basic kinematic equation used in this exercise is: \[ v = u + at \] where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is the acceleration
  • \( t \) is the time in seconds
By applying these equations, we can calculate the velocity of the rocket at the end of its powered phase and determine other significant motion characteristics, such as maximum altitude.
Free Fall
Free fall describes the motion of the rocket when it's solely under the influence of gravity, with no other forces acting on it. Once the rocket's engines stop, and the parachute fails, it enters this stage.During free fall:
  • The only force acting is gravity, which pulls the rocket towards the Earth.
  • The acceleration due to gravity \( g \) is given as \( 32.2 \text{ ft/s}^2 \).
  • Velocity at the peak is zero since it stops moving upwards.
Free fall is characterized by symmetrical movement around the peak altitude, meaning the ascent and descent times are equal. Grasping this concept is essential as it aids in calculating the maximum altitude through time symmetry.
Maximum Altitude Determination
Determining the maximum altitude of the rocket involves understanding both the initial altitude and the final altitude achieved during powered flight and the additional height gained until reaching its peak during free fall.To calculate, we use the formula:\[ h_{additional} = v_1 \cdot t_{up} - \frac{1}{2} \cdot g \cdot t_{up}^2 \]This formula considers:
  • \( v_1 \): the velocity at the end of powered flight.
  • \( t_{up} \): time taken to reach max height upwards.
  • \( g \): gravitational acceleration.
The maximum altitude \( h_{max} \) is then given by adding the additional height attained during free fall ascent to the altitude at the end of powered flight. This systematic approach ensures accurate altitude calculations.
Powered Flight Dynamics
Powered flight dynamics refer to the phase when the rocket's engine is active. The initial speed of the rocket in powered flight is crucial in determining its subsequent motion characteristics. Key Considerations:
  • Determining initial speed at the end of powered flight is vital for subsequent calculations.
  • The acceleration of the rocket propels it upwards until the engine cuts off.
  • The kinematic equations help estimate the speed at this cutoff point using known values like time, acceleration, and end altitude.
Understanding this phase helps us predict how high the rocket will reach before gravity takes over and begins its free fall back to Earth. It's the foundation from which we calculate post-powered motion events.

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Most popular questions from this chapter

A small package is released from rest at \(A\) and moves along the skate wheel conveyor \(A B C D\). The package has a uniform acceleration of \(4.8 \mathrm{m} / \mathrm{s}^{2}\) as it moves down sections \(A B\) and \(C D\), and its velocity is constant between \(B\) and \(C\). If the velocity of the package at \(D\) is \(7.2 \mathrm{m} / \mathrm{s}\), determine \((a)\) the distance \(d\) between \(C\) and \(D,(b)\) the time required for the package to reach \(D .\)

Automobile \(A\) starts from \(O\) and accelerates at the constant rate of \(0.75 \mathrm{m} / \mathrm{s}^{2} .\) A short time later it is passed by bus \(B\) which is traveling in the opposite direction at a constant speed of \(6 \mathrm{m} / \mathrm{s}\). Knowing that bus \(B\) passes point \(O 20 \mathrm{s}\) after automobile \(A\) started from there, determine when and where the vehicles passed each other.

Based on observations, the speed of a jogger can be approximated by the relation \(v=7.5(1-0.04 x)^{0.3}\), where \(v\) and \(x\) are expressed in milh and miles, respectively. Knowing that \(x=0\) at \(t=0\), determine \((a)\) the distance the jogger has run when \(t=1 \mathrm{h},(b)\) the jogger's acceleration in fts at \(t=0,(c)\) the time required for the jogger to run \(6 \mathrm{mi}\).

Car \(A\) is traveling at 40 milh when it enters a 30 milh speed zone. The driver of car \(A\) decelerates at a rate of \(16 \mathrm{ft} / \mathrm{s}^{2}\) until reaching a speed of \(30 \mathrm{mil}\), which she then maintains. When car \(B\), which was initially \(60 \mathrm{ft}\) behind car \(A\) and traveling at a constant speed of \(45 \mathrm{mil}\), enters the speed zone, its driver decelerates at a rate of \(20 \mathrm{ft}\) s " until reaching a speed of \(28 \mathrm{mi} / \mathrm{h}\). Knowing that the driver of car \(B\) maintains a speed of \(28 \mathrm{mi} / \mathrm{h}\), determine \((a)\) the closest that car \(B\) comes to car \(A,(b)\) the time at which car \(A\) is \(70 \mathrm{ft}\) in front of car \(B .\)

During a manufacturing process, a conveyor belt starts from rest and travels a total of \(1.2 \mathrm{ft}\) befors temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to \(\pm 4.8 \mathrm{ft} / \mathrm{s}^{2}\) per second, determine (a) the shortest time required for the belt to move \(1.2 \mathrm{ft}\), (b) the maximum and average values of the velocity of the belt during that time.
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