91Ó°ÊÓ

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) upward. At 25.0 \(\mathrm{s}\) after launch, the rocket fires the second stage, which suddenly boosts its speed to 132.5 \(\mathrm{m} / \mathrm{s}\) upward. This firing uses up all the fuel, however, so then the only force acting on the rocket is gravity. Air resistance is negligible. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

Step by step solution

01

Calculate the height after Stage 1

To find the height reached after the first stage of the rocket, we use the formula for distance under constant acceleration: \[ s = ut + \frac{1}{2}at^2 \]where \(s\) is the final position, \(u\) is the initial speed (0 in this case), \(a\) is the acceleration \(3.50 \, \mathrm{m}/\mathrm{s}^2\), and \(t\) is the time (25.0 \(\mathrm{s}\)).Substituting the known values:\[s = 0 \cdot 25.0 + \frac{1}{2} \times 3.50 \times (25.0)^2 = \frac{1}{2} \times 3.50 \times 625 = 1093.75\, \mathrm{m} \]So, the height after the first stage is 1093.75 meters above the launch pad.

02

Calculate speed just before Stage 2 firing

The speed of the rocket just before the second stage firing can be found using the formula for velocity under constant acceleration:\[v = u + at \]where \(u\) is the initial speed (0 in this case), \(a\) is the acceleration (3.50 \(\mathrm{m}/\mathrm{s}^2\)), and \(t\) is the time (25.0 \(\mathrm{s}\)).Substituting the known values:\[v = 0 + 3.50 \times 25.0 = 87.5 \, \mathrm{m/s}\]This is the speed just before the second stage fires.

03

Calculate initial speed after Stage 2 firing

Immediately after the Stage 2 firing, the speed of the rocket is increased, so the new speed is given:\[ v_{2} = 132.5 \, \mathrm{m/s}\]

04

Calculate maximum height after Stage 2

After Stage 2 fires, the rocket moves upwards until its speed drops to zero due to gravity. Use the formula:\[ v^2 = u^2 + 2as\]where \(v = 0\) at maximum height, \(u = 132.5 \, \mathrm{m/s}\) from Stage 3, \(a = -9.81 \, \mathrm{m/s}^2\).Solving for \(s\):\[0 = (132.5)^2 + 2(-9.81)s\]\[s = -\frac{(132.5)^2}{2(-9.81)} = 894.59 \, \mathrm{m}\]Adding this to the height from stage 1 gives: \[1093.75 + 894.59 = 1988.34 \, \mathrm{m} \]This is the maximum height above the launch pad.

05

Calculate time to fall back to launch pad

To find the total time in the air after Stage 2, add the time to reach the maximum height and the time to fall back.Time to reach maximum height:\[ v = u + at \Rightarrow 0 = 132.5 + (-9.81)t\]\[ t = \frac{132.5}{9.81} = 13.51 \, \mathrm{s}\]Total time to maximum values: 25 s + 13.51 s = 38.51 sTime to fall back: Objects in free fall from height \(h\) under gravity takes\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1988.34}{9.81}}\]\[ t = 20.1 \, \mathrm{s}\]Total time in the air is the sum of ascent and descent times: 13.51 s + 20.1 s = **33.61 s** after stage 2, since the problem asks for time after the second stage firing alone.

06

Calculate speed just before reaching the launch pad

Use the formula for an object in free fall to determine its final speed, given that it falls from rest at the maximum height back to the launch pad:\[v = \sqrt{2gh}\]where \(g = 9.81 \mathrm{m}/\mathrm{s}^2\), \(h = \text{max height from launchpad} = 1988.34\, \mathrm{m}\).\[v = \sqrt{2 \times 9.81 \times 1988.34}\]\[v = 197.4 \, \mathrm{m/s}\]This is the speed just before it hits the launch pad.

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Key Concepts

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

Constant Acceleration

Constant acceleration is a fundamental concept in projectile motion. It refers to a situation where an object's speed changes at a uniform rate over time.
In the exercise, the rocket experiences constant acceleration during its initial phase of flight. Acceleration is measured in meters per second squared (\( \mathrm{m/s}^2 \)), indicating how the velocity increases every second.
Here, the rocket accelerates at \( 3.50 \mathrm{m/s}^2 \) upwards. Understanding constant acceleration involves knowing how to calculate positions and velocities over time:

• Use \( s = ut + \frac{1}{2}at^2 \) for distance covered under constant acceleration.
• The formula \( v = u + at \) helps find the object's velocity.

These formulas help describe how the rocket moves before it shifts to the next phase of its journey.

Initial Velocity

Initial velocity is the speed of an object at the start of an observation.
It's critical in determining the subsequent motion of the object, especially for rockets. When the rocket's second stage fires, its velocity changes suddenly to 132.5\( \mathrm{m/s} \). Understanding initial velocity helps in calculating further steps:

• In our case, initial velocity for stage one is \( 0 \mathrm{m/s} \)
• While for stage two, after the rocket's second boost, it becomes \( 132.5 \mathrm{m/s} \)

This boost determines how high the rocket will finally travel, illustrating how initial conditions set future motion stages.

Maximum Height

The maximum height of a projectile is the highest point it reaches during its motion.
For a rocket, this means the point where it stops rising and begins to fall. To find this height, we need to calculate using the physics of projectile motion. When the second stage fires, the rocket moves upward until gravity slows it to a stop. Here's the process:

• Calculate the height after the first stage using the formula for distance under constant acceleration.
• Add the height gained after the second stage using the velocity at the start of that stage.
• Use \( v^2 = u^2 + 2as \) where \( v = 0 \) at maximum height.

This results in a combined height above the launch pad, which illustrates the complete journey.

Free Fall

Free fall describes the motion of an object under the influence of gravity alone.
Typically, air resistance is ignored, simplifying calculations. After the rocket's second stage, it eventually stops rising and begins to fall back to Earth. This phase is modeled as free fall.

• Time to fall can be calculated using \( t = \sqrt{\frac{2h}{g}} \) with \( g = 9.81 \mathrm{m/s}^2 \).
• The speed just before hitting the ground can be found with \( v = \sqrt{2gh} \).

Understanding free fall helps calculate when the rocket hits the ground, showing its entire time of flight after the second stage's influence ceases.