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Chapter 3: Problem 48

Martian Athietics. In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the greatest horizontal distance. Suppose that on eurth she is in the air for time \(T\) , reaches a maximum height \(h,\) and achieves a horizontal distance \(D\) . If she jumped in exactly the same way during a competition on Mars, where \(g_{\mathrm{Max}}\) is 0.379 of its earth value, find her time in the air, maximum height, and horizontal distance. Express each of these three quantities in terms of its earth value. Air resistance can be neglected on both planets.

Short Answer

Expert verified
Time: \( \frac{T}{\sqrt{0.379}} \), Height: \( \frac{h}{0.379} \), Distance: \( \frac{D}{\sqrt{0.379}} \).

Step by step solution

01

Understanding Earth quantities

On Earth, let the acceleration due to gravity be denoted as \( g \). Thus, the Earth values are: time in the air \( T \), maximum height \( h \), and horizontal distance \( D \). Her vertical motion is influenced by gravity \( g \).
02

Time in the Air on Mars

Since gravity on Mars, denoted as \( g_{\text{Mars}} \), is \( 0.379 \times g \), the time of flight is dependent on the inverse square root of gravity. Thus, the time \( T_{\text{Mars}} \) can be calculated as:\[T_{\text{Mars}} = T \times \sqrt{\frac{g}{g_{\text{Mars}}}} = T \times \sqrt{\frac{1}{0.379}} = \frac{T}{\sqrt{0.379}}\]
03

Calculating Maximum Height on Mars

The maximum height also depends on the inverse of gravity, meaning if gravity decreases, the maximum height increases proportionally. Thus, the maximum height \( h_{\text{Mars}} \) is determined by:\[h_{\text{Mars}} = h \times \frac{g}{g_{\text{Mars}}} = h \times \frac{1}{0.379}\]
04

Horizontal Distance on Mars

The horizontal distance also depends on the time of flight but is unaffected by the vertical acceleration due to gravity during horizontal motion. Thus, the horizontal distance \( D_{\text{Mars}} \) is:\[D_{\text{Mars}} = D \times \frac{T_{\text{Mars}}}{T} = D \times \frac{1}{\sqrt{0.379}}\]
05

Final Expressions

Hence, the athlete's quantities on Mars in terms of Earth's quantities are:- Time in the air: \( T_{\text{Mars}} = \frac{T}{\sqrt{0.379}} \).- Maximum height: \( h_{\text{Mars}} = h \times \frac{1}{0.379} \).- Horizontal distance: \( D_{\text{Mars}} = D \times \frac{1}{\sqrt{0.379}} \).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. In the context of the Martian athletic jump problem, kinematics helps us analyze how the athlete moves through the air. When an athlete jumps, she follows a projectile motion trajectory.
The motion has two main components:
  • Vertical motion: Influenced by gravity, determining how high she goes and how long she stays airborne.
  • Horizontal motion: Not affected by gravity, determining how far she travels horizontally.
During projectile motion, the athlete's motion in the air will be parabolic. The vertical motion is specially influenced by gravity, which plays a substantial role in defining the time she spends in the air and her highest point reached. In contrast, her horizontal motion remains constant since there's no air resistance or horizontal forces acting in this problem.
Gravity on Mars
Gravity on Mars is significantly weaker than on Earth. This difference in gravitational pull arises because Mars has a smaller mass and radius compared to Earth. Specifically, the gravitational acceleration on Mars, denoted as \( g_{\text{Mars}} \), is only about 0.379 times the gravitational pull on Earth, which profoundly affects athlete's jump in various ways.
  • Vertical Motion Alterations: With weaker gravity, Mars allows the athlete to stay in the air longer and reach greater heights. This is due to the inverse relationship between gravitational force and the time of flight, as well as maximum height.
  • Longer Flight Time: For the athlete, less gravitational pull means her time spent in the air is longer, as calculated by \( T_{\text{Mars}} = \frac{T}{\sqrt{0.379}} \).
This weak gravitational influence positively impacts the athlete's performance, providing opportunities to explore longer flight times and higher jumps under the Martian conditions.
Horizontal Distance Calculation
In the scenario of the athlete jumping on Mars, calculating the horizontal distance involves understanding the effect of time in air on horizontal displacement. In projectile motion, when considering the jumping athlete:
  • Time Dependency: The horizontal distance jumped, \( D_{\text{Mars}} \), is directly proportional to the time of flight. Since gravity doesn't influence horizontal motion, although the overall movement is parabolic, the time of flight plays a critical role.
  • Formula Dependence: We calculate the new horizontal range using the time ratio: \( D_{\text{Mars}} = D \times \frac{1}{\sqrt{0.379}} \).
This relationship shows that even though horizontal acceleration is non-existent, the extension in air time due to weaker gravity on Mars results in a longer horizontal distance in comparison to Earth. For students solving this context, understanding that time of flight is crucial helps clarify why and how the horizontal distance is altered.

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Most popular questions from this chapter

A projectile is fired from point \(A\) at an angle above the horizontal. \(\Lambda \mathrm{t}\) its highest point, after having traveled a horizontal distance \(D\) from its launch point, it suddenly explodes into two identical fragments that travel horizontally with equal but opposite velocities as measured relative to the projectile just before it exploded. If one fragment lands back at point \(A,\) how far from A (in terms of \(D )\) does the other fragment land?

Firemen are shooting a stream of water at a burning building using a high- pressure hose that shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 \(\mathrm{s}\) to reach a building 45.0 \(\mathrm{m}\) away. You can ignore air resistance; assume that the end of the hose is at ground level. (a) Find the angle of elevation \(\alpha .\) (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

A squirrel has \(x-\) and \(y\) -coordinates \((1.1 \mathrm{m}, 3.4 \mathrm{m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{m},-0.5 \mathrm{m})\) at time \(t_{2}=3.0 \mathrm{s} .\) For this time interval, find (a) the components of the average velocity, and \((\mathrm{b})\) the magnitude and direction of the average velocity.

A 400.0 -m-wide river flows from west to east at 30.0 \(\mathrm{m} / \mathrm{min}\) . Your boat moves at 100.0 \(\mathrm{m} / \mathrm{min}\) relative to the water no matter which direction you point it. To cross this river, you start from a dock at point \(A\) on the south bank. There is a boat landing directly opposite at point \(B\) on the north bank, and also one at point \(C, 75.0 \mathrm{m}\) downstream from \(B(\text { Fig. } 3.53)\) . (a) Where on the north shore will you land if you point your boat perpendicular to the water current, and what distance will you have traveled? (b) If you initially aim your boat directly toward point \(C\) and do not change that bearing relative to the shore, where on the north shore will you land? (c) To reach point \(C :(i)\) at what bearing must you aim your boat, (ii) how long will it take to cross the river, (iii) what distance do you travel, and (iv) and what is the speed of your boat as measured by an observer standing on the river bank?

A shot putter releases the shot some distance above the level ground with a velocity of \(12.0 \mathrm{m} / \mathrm{s}, 51.0^{\circ}\) above the horizontal. The shot hits the ground 2.08 \(\mathrm{s}\) later. You can ignore air resistance. (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for \(R\) in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.
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