/*! 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 41 A \(7500 \mathrm{~kg}\) rocket b... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(7500 \mathrm{~kg}\) rocket blasts off vertically from the launch pad with a constant upward acceleration of \(2.25 \mathrm{~m} / \mathrm{s}^{2}\) and feels no appreciable air resistance. When it has reached a height of \(525 \mathrm{~m}\), 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
The maximum height above the launch pad the rocket will reach is approximately 971.8 m. After the engine failure, it will take about 44.5 s for the rocket to crash down to the launch pad, and the velocity just before it crashes will be approximately 139.1 m/s.

Step by step solution

01

Calculate time to reach given height

The equation of motion that applies here is \( h = u t + 0.5 a t^{2} \) where \( h = 525m, u = 0 m/s, a = 2.25 m/s^2 \). Solving for \( t \) we get \( t = \sqrt{ \frac{h}{0.5*a} } \) which will result in \( t \approx 19.4s \)
02

Calculate the velocity at that time

Now that we have the time it takes to reach 525m, we can calculate the velocity at that height using the equation \( v = u + at \) where \( u = 0 m/s, a = 2.25 m/s^2, t \approx 19.4s \). This gives us \( v \approx 43.7 m/s \) which is the velocity when the rocket engine fails.
03

Calculate the maximum height the rocket will reach

Now, to calculate the maximum height reached by the rocket. We take into account the rockets upwards velocity when the engines failed and its downward acceleration due to gravity. Here we use the equation of motion \( h_{max} = h + \frac{v^{2}}{2g} \) where \( h = 525m, v \approx 43.7 m/s, g = 9.8 m/s^2 \), we get \( h_{max} \approx 971.8 m \). So, the rocket will reach a maximum height of about 971.8m above the launch pad.
04

Calculate the total time for the rocket to crash to the launch pad

The rocket will return to the ground under the influence of gravity. The time to reach the ground can be calculated by the equation \( t = \sqrt{\frac{2h}{g}} \) where \( h = h_{max} \approx 971.8 m, g = 9.8 m/s^2 \), this give us \( t \approx 44.5s \). Therefore, 44.5s will elapse after engine failure before the rocket comes crashing down to the launch pad.
05

Calculate the final velocity of the rocket

The final velocity of the rocket can be calculated by the equation of motion \( v = \sqrt{2gh} \) where \( h= h_{max} \approx 971.8m, g= 9.8 m/s^2 \). It gives \( v \approx 139.1m/s \). Therefore, the rocket will be moving at 139.1 m/s just before it crashes.
06

Sketching the required diagrams

The acceleration versus time graph ( \(a_{y}-t\) ) will be a straight line at \( 2.25 m/s^2 \) from the start until the engines fail at 19.4s, after then it will stay at \( -9.8 m/s^2 \) due to gravity. The velocity versus time graph ( \(v_{y}-t\) ) will be a straight line rising from 0 m/s at increasing rate of \( 2.25 m/s^2 \) until the engines fail at 19.4s, after which it will decrease at a constant rate of \( 9.8 m/s^2 \) again due to gravity. The displacement versus time graph ( \(y-t\) ) will show a constantly increasing slope from blast-off till the rocket engine fails and then a decelerating downward curve as it falls back to the ground due to gravity. The maximum height reached will be where the slope of the graph is 0 (i.e., at the top of the parabolic curve).

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Key Concepts

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

Kinematic Equations
When studying the motion of objects such as rockets, we rely on the kinematic equations to predict future positions and velocities. These equations describe the motion of an object under constant acceleration, which is quite common, for example, in the case of objects in free fall or, as in our textbook problem, rockets moving under the influence of gravity and thrust.

In the rocket physics problem provided, the kinematic equation used to calculate the time to reach a given height is \( h = ut + \frac{1}{2}at^2 \), where 'h' represents the height (or displacement), 'u' is the initial velocity, 't' is the time, and 'a' is the acceleration. This equation allows us to find the time it took for the rocket to reach 525 meters with an upward acceleration. These equations are pivotal because they form the backbone of understanding motion in a straight line which is particularly useful in a multitude of physics problems including those involving rockets.
Projectile Motion
Projectile motion is a form of motion experienced by an object that is launched into the air and is subject to the forces of gravity and air resistance. For many physics problems, air resistance is considered negligible, and the motion is dictated purely by gravity. This simplification leads to a predictable, parabolic trajectory, which is seen in the textbook problem when the rocket engines fail and the rocket becomes a projectile.

Until engine failure, the rocket's motion could be approached as a simple one-dimensional problem with constant acceleration. After engine failure, however, the rocket's upward motion will slow due to gravity until it stops and begins to fall back to Earth, tracing a symmetric path (in the absence of air resistance). To calculate the maximum height and time to fall back to the launch pad, a basic understanding of projectile motion is crucial, as demonstrated in the solution steps for determining the peak of the trajectory and the duration of the fall.
Free Fall Acceleration
Free fall is the motion of a body where gravity is the only force acting upon it. In our context regarding the rocket problem, after engine failure, the rocket is in free fall. The standard acceleration due to gravity is \( g = 9.8 m/s^2 \) on Earth's surface, and it is this constant acceleration that affects the rocket’s velocity and position over time.

The kinematic equations can be adjusted to include the free fall acceleration when an object is falling downward toward Earth after reaching its peak height — like our rocket. For instance, the equation \( v = \sqrt{2gh} \) can be used to find the velocity of the rocket just before impact. Importantly, regardless of the mass of the object, free fall acceleration remains constant, showcasing one of the fundamental principles of gravity established by Galileo.

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Most popular questions from this chapter

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