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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 second stage fires for \(10.0 \mathrm{~s}\), which boosts the rocket's velocity to \(132.5 \mathrm{~m} / \mathrm{s}\) upward at \(35.0 \mathrm{~s}\) after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stagetwo rocket be moving just as it reaches the launch pad?

Step by step solution

01

Calculation - First Stage Acceleration

Calculate the velocity achieved at the start of the second stage using the equation \(v = u + at\), where \(v\) is the final velocity at the end of the stage, \(u\) is the initial velocity of the stage(=0 m/s as it starts from rest), \(a\) is the acceleration(=3.5 m/s²), and \(t\) is the time taken(=25 s).

02

Calculation - Velocity at Second Stage

Calculate the velocity after the second stage fires. It is gathered from the problem statement that the second stage gives an additional velocity of \(132.5 m/s\), therefore it will need to be added to the velocity obtained at the end of the first stage.

03

Calculation - Maximum Height

Calculate the maximum height using equation \(v^2 = u^2 + 2gh\), where \(v\) is the final velocity after fuel is exhausted(= 0 m/s), \(u\) is the initial velocity(= velocity calculated at end of second stage), \(g\) is the acceleration due to gravity(= -9.8 m/s² since it acts downward), and \(h\) is the height we need to find.

04

Calculation - Time to fall back to ground

Knowing that the object falls from the calculated height \(h\), use equation \(h = ut + 1/2 * g * t^2\) where \(u\) is the initial velocity(= 0 m/s), \(g\) is acceleration due to gravity (= 9.8 m/s² since it is downward), and \(t\) is the time to fall back (solved for).

05

Calculation - Velocity before hitting the ground

Using the same equation of motion \(v = u + gt\) where \(v\) is the velocity just before it hits the ground, \(u\) is the initial velocity (= 0 m/s) and \(g\) is the acceleration due to gravity (= 9.8 m/s²), one can calculate for \(v\) with the time obtained from previous step.

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Key Concepts

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

Constant Acceleration

When we talk about constant acceleration in physics, it means that the velocity of an object increases by the same amount each second. In our two-stage rocket problem, the first stage involves constant acceleration. The rocket starts at rest and accelerates upwards with a constant acceleration of \(3.5 \text{ m/s}^2\).

• Because the acceleration is constant, we can use the formula \( v = u + at \) to find how its velocity changes over time.
• Here, \( u \) is the initial velocity, which is \(0 \text{ m/s}\) because the rocket starts from a complete stop.


• \( a \) is the acceleration, which remains at \(3.5 \text{ m/s}^2\).
• \( t \) is the time in seconds, which is \(25 \text{ s}\) for the first stage.

This simple equation allows us to calculate the velocity at the end of the first stage. Constant acceleration simplifies our calculations, as the velocity change over time is predictable and straightforward.

Projectile Motion

Projectile motion occurs when an object is thrown into space and affected only by gravity once the initial propulsion stops. In the case of the rocket, after both stages have fired and the fuel is exhausted, it becomes a projectile.

• For the rocket at the end of the second stage, its motion can be analyzed using the concepts of projectile motion.
• Initially, it has an upward velocity which will gradually decrease as it moves upward, due to the work of gravity acting downwards.


• The peak of its flight, or maximum height, occurs when its vertical velocity reaches zero.
• At this point, the only force acting on it is gravity.

Using the equation \( v^2 = u^2 + 2gh \), where the final velocity \( v \) is zero at maximum height, and \( u \) is the upward velocity after stage two completes firing, we can calculate the maximum height it achieves.

Gravity Effects

Gravity is a constant force that acts on all objects with mass near Earth's surface. It pulls objects toward the Earth at \(9.8 \text{ m/s}^2\). For this rocket, once the fuel is exhausted, gravity becomes the dominating force.

• Gravity decelerates the rocket as it travels upward, eventually bringing it to a stop at its maximum height.


• Once the rocket reaches this apex, gravity then accelerates it downward.
• The time it takes to fall back to the launch pad can be found using the formula \( h = ut + \frac{1}{2} gt^2 \).
• "\( g \) is gravity’s acceleration (\(9.8 \text{ m/s}^2\)), and \( t \) can be solved for the descent.

At the end of its fall, the speed at which the rocket hits the ground can be calculated with \( v = u + gt \). Here, \( u = 0 \text{ m/s} \) just as it begins to fall back."
Understanding gravity’s role is crucial in predicting how the rocket will move after it runs out of fuel, ensuring a smooth analysis of its motion.