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\(\mathrm{A}\) certain volcano on earth can eject rocks vertically to a maximum height \(H .\) (a) How high (in terms of \(H\) ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is \(3.71 \mathrm{m} / \mathrm{s}^{2},\) and you can neglect air resistance on both planets. (b) If the rocks are in the air for a time \(T\) on earth, for how long (in terms of \(T\)) will they be in the air on Mars?

Short Answer

Expert verified
(a) The height on Mars is \( 2.644H \). (b) The time on Mars is \( 2.644T \).

Step by step solution

01

Determine the relationship for maximum height

The maximum height reached by a projectile is given by the formula \( H = \frac{v^2}{2g} \), where \( v \) is the initial velocity, and \( g \) is the acceleration due to gravity. On Earth, the height is given by \( H = \frac{v^2}{2g_\text{Earth}} \). We need to find the corresponding height \( H_\text{Mars} \) on Mars, using \( g_\text{Mars} \).
02

Express height on Mars in terms of height on Earth

On Mars, the maximum height is \( H_\text{Mars} = \frac{v^2}{2g_\text{Mars}} \). From the relationship \( H = \frac{v^2}{2g} \), we can express this as \( H_\text{Mars} = \left(\frac{g_\text{Earth}}{g_\text{Mars}}\right)H \). Substituting \( g_\text{Earth} = 9.81 \, \mathrm{m/s^2} \) and \( g_\text{Mars} = 3.71 \, \mathrm{m/s^2} \), we find \( H_\text{Mars} = \left(\frac{9.81}{3.71}\right)H \).
03

Simplify the expression for height on Mars

Calculate the ratio \( \frac{9.81}{3.71} \) to obtain the expression \( H_\text{Mars} = 2.644H \). This means that the rocks would reach a height of \( 2.644H \) on Mars.
04

Determine the time of flight relationship

The time of flight for an object projected vertically is given by \( T = \frac{2v}{g} \). On Earth, \( T_\text{Earth} = \frac{2v}{g_\text{Earth}} \) and on Mars, \( T_\text{Mars} = \frac{2v}{g_\text{Mars}} \).
05

Express time of flight on Mars in terms of Earth time

Using \( T_\text{Mars} = \frac{2v}{g_\text{Mars}} \) and the Earth equivalent, we can express the Mars time in terms of Earth time: \( T_\text{Mars} = \left(\frac{g_\text{Earth}}{g_\text{Mars}}\right)T_\text{Earth} \). Substituting the values, we get \( T_\text{Mars} = 2.644T \).
06

Simplify the expression for time on Mars

The derived expression \( T_\text{Mars} = 2.644T \) indicates that the rocks would be in the air on Mars for a time \( 2.644T \).

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Key Concepts

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

Acceleration Due to Gravity
When we talk about projectile motion, the term 'acceleration due to gravity' is fundamental. This refers to the force that pulls objects towards the center of a planet. Every planetary body, like Earth or Mars, has its own unique gravitational acceleration.

- On Earth, the acceleration due to gravity is approximately 9.81 meters per second squared \( (m/s^2) \).
- On Mars, it's much less, around 3.71 \( m/s^2 \).

This difference plays a crucial role in determining how high a projectile can go and how long it will stay in the air. A lesser gravitational force on Mars means that, with the same initial force, a projectile can reach further heights compared to Earth. This is because gravity impacts the projectile less intensely. The formula to calculate the maximum height or time of flight involves these gravitational values, hence highlighting their importance.
Maximum Height
The maximum height that a projectile can achieve is a critical aspect of its journey. It tells us how high the object will fly before starting its descent back down. The formula to calculate this height is given by:\[ H = \frac{v^2}{2g} \]

Here, \( H \) is the maximum height, \( v \) is the initial velocity, and \( g \) is the acceleration due to gravity. This formula shows that as gravity decreases, the same initial velocity takes the projectile to a greater height.

- On Earth, with a higher \( g \), the height is less.
- On Mars, with its lower \( g \), the same velocity results in a larger maximum height.

In the given exercise, by substituting the respective gravity values of Earth and Mars, we find that rocks ejected from Mars reach a height 2.644 times higher than those from Earth, due to the difference in gravitational accelerations.
Time of Flight
Time of flight refers to the duration a projectile remains in the air before returning to the ground. This time is influenced by the initial velocity and the acceleration due to gravity. It can be calculated using the equation:\[ T = \frac{2v}{g} \]

- On Earth, this results in a shorter time compared to Mars.
- On Mars, despite having the same initial velocity, the rocket stays airborne longer.

For this exercise, by replacing the gravitational accelerations, the projectile on Mars stays 2.644 times longer in the air than on Earth. Less gravitational pull allows it to remain airborne longer, similar to how altitude increases. This relationship helps in planning trajectories in different gravitational fields, impacting fields like space exploration and even sports science.

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Most popular questions from this chapter

A Multistage Rocket. 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 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 s after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance can be neglected. (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 stage-two rocket be moving just as it reaches the launch pad?

The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the \(x\)-axis). \(\begin{array}{llll}{\text { Time }(s)} & {0} & {2.1} & {20.0} & {53} \\\ {\text { Speed (mijh) }} & {0} & {60} & {200} & {253}\end{array}\) (a) Make a \(v_{x}-t\) graph of this car's velocity \((\) in \(\mathrm{mi} / \mathrm{h})\) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration ( in \(\mathrm{m} / \mathrm{s}^{2} )\) between ( i 0 and \(2.1 \mathrm{s} ;\) (ii) 2.1 \(\mathrm{s}\) and 20.0 \(\mathrm{s}\); (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs \(\$ 1.25\) million!)

Passing. The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) (about 45 \(\mathrm{mi} / \mathrm{h} )\) . Initially, the car is also traveling at 20.0 \(\mathrm{m} / \mathrm{s}\) and its front bumper is 24.0 \(\mathrm{m}\) behind the truck's rear bumper. The car accelerates at a constant 0.600 \(\mathrm{m} / \mathrm{s}^{2}\) , then pulls back into the truck's lane when the rear of the car is 26.0 \(\mathrm{m}\) ahead of the front of the truck. The car is 4.5 \(\mathrm{m}\) long and the truck is 21.0 \(\mathrm{m}\) long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where acceleration of the rocket is given by \(a_{y}=\left(2.80 \mathrm{m} / \mathrm{s}^{3}\right) t,\) where the \(+y\) -direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t=10.0 \mathrm{s} ?\) (b) What is the speed of the rocket when it is 325 \(\mathrm{m}\) above the surface of the earth?

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 2.50 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick's velocity just before it reaches the ground? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the brick.

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