/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A golf ball struck on earth rise... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 38

A golf ball struck on earth rises to a maximum height of 60 m and hits the ground 230 m away. How high will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth.

Short Answer

Expert verified
The golf ball will reach a height of approximately 363.33 meters on the moon.

Step by step solution

01

Understand the Problem

We need to find the maximum height a golf ball reaches on the moon, given it rises to 60 m on Earth. The golf ball trajectory is a parabola on both Earth and the Moon. Earth's acceleration due to gravity is approximately 9.81 m/s虏, and on the moon, it is 0.165 times that, which is around 1.62 m/s虏.
02

Equate Maximum Height Formulas

The maximum height of a projectile is given by the formula \( h = \frac{v^2 \sin^2 \theta}{2g} \), where \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity. For both Earth and Moon, \( v \sin \theta \) is constant as the initial conditions and angles are the same.
03

Determine Moon's Height Formula

Since the trajectory is parabolic and initial conditions are the same, draw upon the relationship that maximum height is inversely proportional to gravity \( (h \propto \frac{1}{g}) \). On Earth, \(h_{earth} = 60\) m and \(g_{earth} = 9.81\) m/s虏. On Moon, \(g_{moon} = 0.165 \times g_{earth} = 1.62\) m/s虏.
04

Calculate Maximum Height on Moon

Using proportionality, \( h_{moon} = h_{earth} \times \frac{g_{earth}}{g_{moon}} \). Substitute the values: \( h_{moon} = 60 \times \frac{9.81}{1.62} \).
05

Solve for Maximum Height on the Moon

Calculating the above expression gives \( h_{moon} = 60 \times 6.0556 \approx 363.33 \) m. So, the maximum height on the moon is approximately 363.33 meters.

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Key Concepts

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

Projectile Trajectory
When a golf ball is hit, it follows a curved path called a projectile trajectory. This path is determined by the initial speed, angle of launch, and the gravitational pull acting on the ball. Projectile trajectory is vital in physics to understand how objects move in different environments, such as Earth or the Moon.

Key elements influencing projectile motion include:
  • Initial velocity: The speed at which the ball starts its journey.
  • Launch angle: The angle at which the projectile is launched affects the shape of its trajectory.
  • Gravity: On Earth, gravity pulls the ball downward at 9.81 m/s虏, while on the Moon, it is significantly less at 1.62 m/s虏.
Understanding these factors allows us to predict how far and high a projectile will travel.
Parabolic Motion
The trajectory of a projectile is typically parabolic in shape. This means it is a symmetrical curve, like an arch. The highest point of this arch is known as the vertex.

Parabolic motion comes from the constant acceleration of gravity acting on the projectile, contrasting with the initial horizontal and vertical speeds. The mathematical equation for a parabola helps calculate important aspects like maximum height and range.

In this golf ball example, the parabolic motion remains the same on the Moon as it is on Earth, only the scale changes due to lower gravity.
Gravitational Acceleration
Gravity is a crucial force that affects all projectiles. Gravitational acceleration on Earth is approximately 9.81 m/s虏. However, this value changes depending on the celestial body. For instance, on the Moon, gravity is only about 0.165 of Earth's, making it 1.62 m/s虏.

This difference in gravity greatly impacts how projectiles behave. With less gravity on the Moon, objects travel higher and farther than they would on Earth for the same initial conditions.

Grasping gravitational differences helps us plan and predict scenarios in varying environments like space exploration.
Physics Problem Solving
Tackling physics problems often involves several steps:
  • Identify known and unknown variables: This helps frame the problem.
  • Apply appropriate physics laws or formulas: Use mathematical equations to relate the known to the unknown.
  • Solve with careful calculations: Use algebra and numerical computation to find answers.
  • Verify results: Double-check solutions for accuracy and plausibility.
In the example problem, we identified key data, such as heights and gravity acceleration, applied the formula for maximum height, and performed calculations to solve how high the ball would go on the Moon. Following a structured problem-solving approach ensures accurate and reliable results.

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Most popular questions from this chapter

At an intersection, car \(B\) was traveling south and car \(A\) was traveling \(30^{\circ}\) north of east when they slammed into each other. Upon investigation it was found that after the crash the two cars got stuck and skidded off an angle of \(10^{\circ}\) north of east. Each driver claimed that he was going at the speed limit of \(50 \mathrm{km} / \mathrm{h}\) and that he tried to slow down but couldn鈥檛 avoid the crash because the other driver was going a lot faster. Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively, determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit.

A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supported by springs. Knowing that the height of the rebound is 0.6 m, determine (a) the velocity of the plate immediately after the impact, (b) the energy lost due to the impact.

Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. The train is traveling down a 5-percent grade when it decreases its speed at a constant rate from 120 mi/h to 60 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the baggage car if the trunk is not to slide.

A \(1400-\mathrm{kg}\) automobile starts from rest and travels \(400 \mathrm{m}\) during a performance test. The motion of the automoile is defined by the relation \(a=3.6 e^{-0.0005 x}\), where \(a\) and \(x\) are expressed in \(\mathrm{m} / \mathrm{s}^{2}\) and meters, respectively. The magnitude of the aerodynamic drag is \(D=0.35 v^{2},\) where \(D\) and \(v\) are expressed in newtons and \(\mathrm{m} / \mathrm{s}\), respectively. Determine the power dissipated by the aerodynamic drag when ( \(a) x=200 \mathrm{m},(b) x=400 \mathrm{m} .\)

At an amusement park there 200 -kg bumper cars \(A, B\), and \(C\) that have riders with masses of \(40 \mathrm{kg}, 60 \mathrm{kg}\), and \(35 \mathrm{kg}\), respectively. Car \(A\) is moving to the right with a velocity \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{m} / \mathrm{s}\) and \(\mathrm{car} C\) has a velocity \(\mathrm{v}_{B}=1.5 \mathrm{m} / \mathrm{s}\) to the left, but car \(B\) is initially at rest. The coefficient of restitution between each car is \(0.8 .\) Determine the final velocity of each car, after all impacts, assuming (a) cars \(A\) and \(C\) hit car \(B\) at the same time, \((b)\) car \(A\) hits car \(B\) before car \(C\) does.
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