/*! 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 43 A 7500-kg rocket blasts off vert... [FREE SOLUTION] | 91Ó°ÊÓ

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A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s\(^2\) and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

Short Answer

Expert verified
(a) Max height is 645.6 m. (b) The rocket takes 9.92 s to hit the ground and is moving at 101.4 m/s. (c) Graphs are sketched as described.

Step by step solution

01

Calculate Velocity at Engine Failure

First, determine the velocity of the rocket at 525 meters height before the engines fail. Use the kinematic equation\[ v^2 = u^2 + 2as \]where \( u = 0 \) m/s (initial velocity), \( a = 2.25 \) m/s\(^2\) (acceleration), and \( s = 525 \) m (displacement).Substitute these numbers:\[ v^2 = 0 + 2 \times 2.25 \times 525 \]\[ v^2 = 2362.5 \]\[ v = \sqrt{2362.5} \approx 48.6 \text{ m/s} \]
02

Calculate Additional Height After Engine Failure

Now, calculate how much higher the rocket will travel after engine failure using\[ v^2 = u^2 + 2as \]where \( v = 0 \) m/s (velocity at the maximum height), \( u = 48.6 \) m/s from Step 1, and \( a = -9.8 \) m/s\(^2\) (acceleration due to gravity).\[ 0 = 48.6^2 + 2(-9.8)s \]\[ 0 = 2362.5 - 19.6s \]\[ s = \frac{2362.5}{19.6} \approx 120.6 \text{ m} \]
03

Calculate Maximum Height Above Launch Pad

Add the height reached before and after engine failure to get the maximum height:\[ 525 \text{ m} + 120.6 \text{ m} = 645.6 \text{ m} \]
04

Calculate Time to Fall Back Post-Engine Failure

Using the equation of motion\[ v = u + at \]for the downward journey from maximum height, where \( v = 0 \) m/s (initial velocity at maximum height), \( u = -48.6 \) m/s (velocity just before it falls back), and \( a = 9.8 \) m/s\(^2\):\[ 0 = -48.6 + 9.8t \]\[ 9.8t = 48.6 \]\[ t = \frac{48.6}{9.8} \approx 4.96 \text{ s} \] (time to drop to height of 525 m).Next, calculate the time from 525 m height to the launch pad using same formula where initial velocity \( u = 0 \) m/s, \( v = 48.6 \) m/s and \( a = 9.8 \) m/s\(^2\):\[ 48.6 = 0 + 9.8t \]\[ t = \frac{48.6}{9.8} \approx 4.96 \text{ s} \]Total time elapsed after engine failure = \( 4.96 + 4.96 \approx 9.92 \text{ s} \)
05

Calculate Impact Velocity on Return

The velocity just before hitting the ground can be obtained using:\[ v^2 = u^2 + 2as \]where \( u = 0 \) m/s, \( a = 9.8 \) m/s\(^2\), and \( s = 525 \) m (from height back to launch pad).\[ v^2 = 0 + 2 \times 9.8 \times 525 \]\[ v^2 = 10290 \]\[ v = \sqrt{10290} \approx 101.4 \text{ m/s} \]
06

Sketch the Graphs

1. **Acceleration-Time Graph \( (a_y-t) \):** - From launch until engine failure (0 to some time \(t\)), acceleration is constant at 2.25 m/s\(^2\). - After engine failure, acceleration is constant at -9.8 m/s\(^2\).2. **Velocity-Time Graph \( (v_y-t) \):** - Velocity increases steadily with 2.25 m/s\(^2\) until engine failure; then linearly decreases at rate of 9.8 m/s\(^2\) until it reaches zero (maximum height). - It then increases negatively until it impacts the ground.3. **Position-Time Graph \( (y-t) \):** - Starts from origin, moves upward (concave down) until maximum height, then comes downwards symmetrically to its starting point (the launch pad).

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause it. In this context, we focus on the motion of a rocket. When studying kinematics, we use a set of equations called kinematic equations to solve for variables like velocity, acceleration, displacement, and time.
One of the main kinematic equations used here is:
  • \[ v^2 = u^2 + 2as \], where v is the final velocity, u is the initial velocity, a is acceleration, and s is displacement.
This equation helps us find the velocity of the rocket when the engines fail and the additional height reached after engine failure.
Understanding kinematics helps us predict and describe the motion of objects systematically. By applying this to the problem, we understand the rocket's behavior in both upward motion (considering engine operation) and downward motion (after engine failure).
Acceleration
Acceleration is the rate of change of velocity with time. It tells us how quickly an object speeds up or slows down. In our problem, there are two main phases of acceleration:
  • While the rocket's engines are working, it accelerates upward at 2.25 m/s decreedMath^2, meaning its velocity increases by 2.25 m/s every second.
  • Once the engines fail, the only force acting is gravity, causing an acceleration of -9.8 m/decreedMath^2 (downward).
Understanding acceleration is crucial in physics problem solving. It enables us to predict how an object's speed and direction will change, helping in calculating how far and how fast the rocket moves after engine cutoff.
Velocity
Velocity is a vector quantity that indicates how fast something is moving in a particular direction. It is crucial in the rocket problem because:
  • Initial velocity is zero when the rocket starts.
  • At engine failure, we calculate the velocity using the kinematic equation (with acceleration and height known).
  • Knowing velocity is necessary to determine how far the rocket will travel after the engines stop and how fast it will impact the ground.
Velocity isn't just about speed; it gives us direction, too! Here, the change in velocity guides us through the rocket's journey from the launch pad to the highest point and back.
Gravity
Gravity is a constant force that pulls objects toward the Earth's center at 9.8 m/s decreedMath^2. During the rocket's flight, gravity affects it in numerous ways:
  • When the engines fail, gravity becomes the only force acting on the rocket, pulling it back toward Earth.
  • Gravity's consistent downward acceleration helps determine the rocket's trajectory and movement after engine failure.
By understanding gravity, we can predict both the maximum height after engine cutoff and the behavior of the rocket as it returns to the ground. It's an invisible force but incredibly vital for solving physics problems involving motion.
Physics Problem Solving
Physics problem solving involves systematically approaching complex situations to find solutions. Here’s how this applies to our rocket exercise:
  • Breakdown the motion into stages (powered and non-powered phases).
  • Use relevant physics formulae, like kinematics equations, for each stage.
  • Calculate critical values like velocities at different points, maximum heights, and impacts.
  • Visualize the problem with sketches of graphs (acceleration, velocity, and position over time).
By employing these physics problem-solving strategies, we find logical and mathematical solutions to the rocket's journey, making concepts clearer and more understandable to students.

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