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Chapter 8: Problem 8

A bottle rocket is shot straight up in the air with a speed \(30 \mathrm{m} / \mathrm{s} .\) If the air resistance is ignored, the bottle would go up to a height of approximately \(46 \mathrm{m}\). However, the rocket goes up to only \(35 \mathrm{m}\) before returning to the ground. What happened? Explain, giving only a qualitative response.

Short Answer

Expert verified
The discrepancy in height values (35 m instead of 46 m) can be explained by the presence of air resistance. Air resistance opposes the rocket's upward motion, causing it to lose some of its initial kinetic energy as it goes up, resulting in a lower maximum height than it would reach without air resistance.

Step by step solution

01

If there were no air resistance, the only force acting on the rocket would be gravity. In this case, the motion could be characterized by the following kinematic equation for the height (displacement) of the rocket: \(h = v_0t - \frac{1}{2}gt^2\) where \(h\) is the height, \(v_0\) is the initial velocity (30 m/s), \(t\) is the time taken to reach the highest point, and \(g\) is the acceleration due to gravity (approximately 9.81 m/s虏). At the highest point, the vertical velocity of the rocket is 0, corresponding to a time \(t\) for which we can find the theoretical height without air resistance. #Step 2: Calculate the theoretical height without air resistance#

At the highest point, the vertical velocity is zero (\(v = 0\)). Using the equation \(v = v_0 - gt\), where \(v_0\) is 30 m/s, and \(g\) is 9.81 m/s虏, we can solve for the time \(t\) when the rocket reaches the peak: \(0 = 30 - (9.81)t\) \(t \approx 3.06\,s\) Now we can substitute the calculated time value into the height equation: \(h = 30(3.06) - \frac{1}{2}(9.81)(3.06)^2\) \(h \approx 46\,m\) The rocket would go up to a height of approximately 46 meters if there was no air resistance. #Step 3: Explain the role of air resistance in the rocket's motion#
02

In reality, the rocket is subject to air resistance, a force that opposes its upward motion. Due to air resistance, the rocket loses some of its initial kinetic energy as it goes up, which causes it to reach a lower maximum height than it would without air resistance. When air resistance is taken into account, the actual height reached (35 m) is less than the theoretical height without air resistance (46 m). #Step 4: Qualitative response#

To answer the question qualitatively, we can state that the reason the rocket goes up to only 35 meters instead of 46 meters is due to the presence of air resistance. Air resistance acts against the rocket's upward motion, causing it to lose energy and reach a lower maximum height than it would if there was no air resistance.

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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 the cornerstone of classical mechanics, offering essential tools for predicting an object's position and velocity at any given time. These equations derive from Newton's laws of motion and allow us to analyze the motion of objects when the acceleration is uniform.

One of the key kinematic equations for projectile motion is \(h = v_0t - \frac{1}{2}gt^2\), where \(h\) represents the displacement, \(v_0\) is the initial velocity, \(t\) is time, and \(g\) is the acceleration due to gravity, a constant value of approximately 9.81 \(\mathrm{m/s}^2\) on Earth. When a bottle rocket is launched upwards, this equation can predict its maximum height without air resistance.
Projectile Motion
Projectile motion describes the motion of an object thrown or launched into the air and subject to only the force of gravity. This two-dimensional motion can be split into horizontal and vertical components, but in the case of the bottle rocket shot straight up, we'll focus on the vertical component.

In ideal conditions, with no air resistance, a projectile's vertical motion is symmetrical. The time it takes to rise to its peak is equal to the time it takes to fall back to its original position. The kinematic equations for projectile motion enable us to predict the behavior of the rocket throughout its flight, provided we ignore air resistance. However, as seen with our bottle rocket, real-life conditions with air resistance disrupt this symmetry and affect the trajectory and maximum height achieved.
Acceleration Due to Gravity
The acceleration due to gravity, denoted by \(g\), is the rate at which objects accelerate towards the Earth when they are in free fall. Its standard value is 9.81 \(\mathrm{m/s}^2\) and plays a vital role in kinematic equations. Gravity is the primary force considered in projectile motion calculations when air resistance is neglected.

For instance, when we calculate that the bottle rocket should reach a height of 46 meters, this assumes that gravity is the sole force acting upon it. This constant acceleration downwards is why the rocket eventually stops ascending and begins its descent, even in an ideal scenario without air resistance.
Energy Loss
One of the realities of projectile motion in a non-vacuum environment like Earth's atmosphere is energy loss due to air resistance. As the rocket ascends, it collides with air molecules, transferring some kinetic energy from the rocket to the air, effectively slowing it down.

This energy loss is the reason the actual height reached by the bottle rocket, 35 meters, is less than the theoretical height calculated without considering air resistance. Energy loss to air resistance decreases the rocket's vertical component of velocity more quickly than gravity alone would, leading to a lower peak before gravity pulls the rocket back to the ground.

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Most popular questions from this chapter

A force \(F(x)=(3.0 / x) \mathrm{N}\) acts on a particle as it moves along the positive \(x\) -axis. (a) How much work does the force do on the particle as it moves from \(x=2.0 \mathrm{m}\) to \(x=5.0 \mathrm{m} ?\) (b) Picking a convenient reference point of the potential energy to be zero at \(x=\infty,\) find the potential energy for this force.

A child jumps up and down on a bed, reaching a higher height after each bounce. Explain how the child can increase his maximum gravitational potential energy with each bounce.

A camera weighing \(10 \mathrm{N}\) falls from a small drone hovering \(20 \mathrm{m}\) overhead and enters free fall. What is the gravitational potential energy change of the camera from the drone to the ground if you take a reference point of (a) the ground being zero gravitational potential energy? (b) The drone being zero gravitational potential energy? What is the gravitational potential energy of the camera (c) before it falls from the drone and (d) after the camera lands on the ground if the reference point of zero gravitational potential energy is taken to be a second person looking out of a building \(30 \mathrm{m}\) from the ground?

A cat's crinkle ball toy of mass \(15 \mathrm{g}\) is thrown straight up with an initial speed of \(3 \mathrm{m} / \mathrm{s}\). Assume in this problem that air drag is negligible. (a) What is the kinetic energy of the ball as it leaves the hand? (b) How much work is done by the gravitational force during the ball's rise to its peak? (c) What is the change in the gravitational potential energy of the ball during the rise to its peak? (d) If the gravitational potential energy is taken to be zero at the point where it leaves your hand, what is the gravitational potential energy when it reaches the maximum height? (e) What if the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? (f) What is the maximum height the ball reaches?

A block of mass \(m\), after sliding down a frictionless incline, strikes another block of mass \(M\) that is attached to a spring of spring constant \(k\) (see below). The blocks stick together upon impact and travel together. (a) Find the compression of the spring in terms of \(m, M, h, g,\) and \(k\) when the combination comes to rest. Hint: The speed of the combined blocks \(m+M\left(v_{2}\right)\) is based on the speed of block \(m\) just prior to the collision with the block \(M\left(\mathrm{v}_{1}\right)\) based on the equation \(v_{2}=(m / m)+M\left(v_{1}\right) .\) This will be discussed further in the chapter on Linear Momentum and Collisions. (b) The loss of kinetic energy as a result of the bonding of the two masses upon impact is stored in the so-called binding energy of the two masses. Calculate the binding energy.
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