/*! 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 30 Estimate how much farther a pers... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimate how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.

Short Answer

Expert verified
A person can jump six times farther on the Moon than on Earth.

Step by step solution

01

Understand the Problem

We need to determine the difference in jump distance on the Moon compared to Earth, given the same takeoff speed and angle. The gravity on the Moon is one-sixth of that on Earth.
02

Recall the Projectile Motion Formula

The horizontal range of a projectile is given by the formula \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( v_0 \) is the initial speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.
03

Express Earth and Moon Gravity

Let \( g_E \) be the gravity on Earth and \( g_M = \frac{1}{6}g_E \) be the gravity on the Moon.
04

Calculate Distance on Earth

The distance a person jumps on Earth is \( R_E = \frac{v_0^2 \sin(2\theta)}{g_E} \).
05

Calculate Distance on Moon

The distance a person jumps on the Moon is \( R_M = \frac{v_0^2 \sin(2\theta)}{g_M} = \frac{v_0^2 \sin(2\theta)}{\frac{1}{6}g_E} = 6 \times \frac{v_0^2 \sin(2\theta)}{g_E} \).
06

Compare Distances

By comparing \( R_M \) and \( R_E \), we find \( R_M = 6 R_E \). Therefore, the jump distance on the Moon is six times that on Earth.

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Key Concepts

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

Gravity on the Moon
Gravity on the Moon is a crucial factor affecting how motion and activities differ from those on Earth. While gravity is a force pulling objects toward the center of a celestial body, its strength can vary significantly based on that body's mass and size. The Moon, being much smaller than Earth, has a gravitational strength that is only one-sixth of Earth’s gravity. This reduced gravity means that everything from jumping to walking might seem much easier and feel lighter.

On Earth, the average acceleration due to gravity is approximately 9.81 meters per second squared (\( 9.81 \, m/s^2 \)). On the Moon, however, this figure drops significantly to about 1.63 meters per second squared (\( 1.63 \, m/s^2 \)). Understanding the extent of this reduction is essential for calculating how various activities and motions, such as jumping, will be affected. Knowing that gravity on the Moon is weaker helps us anticipate the increased distances and times associated with moving objects on its surface.
Jump Distance
The distance one can jump is substantially affected when moving from Earth's gravity to the Moon's lower gravitational force. By analyzing projectile motion, we can see how a jump's horizontal range increases under reduced gravity.

When a person jumps on Earth, the range or distance traveled is determined by their initial speed, the angle of takeoff, and the gravity pulling them down. Using the formula for projectile motion's horizontal range, which is \[ R = \frac{v_0^2 \sin(2\theta)}{g} \], the acceleration due to gravity directly influences how far the jump will go. The jump distance on Earth is limited by the stronger gravitational pull \( (g_E) \).

On the Moon, with a gravity that is one-sixth of Earth’s, the formula adapts accordingly. Here, the weaker pull, represented as \( g_M = \frac{1}{6} g_E \), allows for a longer trajectory. Substituting Moon's gravity into the range equation significantly extends the jump distance, enabling jumps that are six times farther than on Earth, assuming takeoff conditions like speed and angle are unchanged.
Acceleration due to Gravity
The concept of acceleration due to gravity is fundamental in understanding how objects move on different planets or celestial bodies. It refers to the rate at which velocity changes because of gravity's pull. On Earth, we experience a certain rate of acceleration (\( 9.81 \, m/s^2 \)), which heavily influences all our physical activities.

When considering the Moon, the acceleration due to gravity is dramatically reduced to \( 1.63 \, m/s^2 \).This smaller rate of acceleration means that objects fall slower and move more sluggishly when pushed or thrown. Without Earth's gravitational pull to heavily influence motion, activities on the Moon become vastly different.

This contrast impacts a variety of scenarios, from how astronauts maneuver through lunar missions to what it would feel like to jump or run. With reduced acceleration due to gravity, objects and people on the Moon encounter less resistance in their motion, leading to expanded ranges and longer airtime in projectile motions, like jumps.

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Most popular questions from this chapter

(II) A shot-putter throws the shot (mass \(=7.3 \mathrm{kg}\) ) with an initial spced of 14.4 \(\mathrm{m} / \mathrm{s}\) at a \(34.0^{\circ}\) angle to the borizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 \(\mathrm{m}\) above the ground.

(II) A ball is thrown horizontally from the roof of a building 9.0 \(\mathrm{m}\) tall and lands 9.5 \(\mathrm{m}\) from the base. What was the ball's initial spced?

Two planes approach each other head-on. Each has a speed of \(780 \mathrm{~km} / \mathrm{h}\), and they spot each other when they are initially \(12.0 \mathrm{~km}\) apart. How much time do the pilots have to take evasive action?

Huck Finn walks at a speed of \(0.70 \mathrm{~m} / \mathrm{s}\) across his raft (that is, he walks perpendicular to the raft's motion relative to the shore). The raft is traveling down the Mississippi River at a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to the river bank (Fig. \(3-49\) ). What is Huck's velocity (speed and direction) relative to the river bank?

(11) (a) A long jumper leaves the ground at \(45^{\circ}\) above the horizontal and lands 8.0 \(\mathrm{m}\) away. What is her "takeoff" spced \(v_{0} ?(b)\) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 \(\mathrm{m}\) away horizontally and \(2.5 \mathrm{m},\) vertically below. If she long jumps from the edge of the left bank at \(45^{\circ}\) with the speed calculated in \((a),\) how long, or short, of the opposite bank will she land (Fig. 43\() ?\)

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