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(II) 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

Analyze the Problem

First, identify the given information: the takeoff speed and angle are the same on both the Moon and the Earth. The gravity on the Moon is one-sixth of Earth's gravity, which is \( g_m = \frac{1}{6}g_e \). We need to find how much farther a person can jump on the Moon.
02

Formula for Range of a Projectile

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

Calculate Range on Earth and Moon

Compute the range for both Moon and Earth:For Earth, use:\[ R_e = \frac{v^2 \sin(2\theta)}{g_e} \]For the Moon, since gravity is one-sixth of Earth's, substitute:\[ R_m = \frac{v^2 \sin(2\theta)}{\frac{1}{6}g_e} = 6\cdot \frac{v^2 \sin(2\theta)}{g_e}\]
04

Compare the Two Ranges

Compare the formulas derived:\[ R_m = 6 \cdot R_e \]This shows that the range 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
Gravity is a fundamental force that attracts two bodies towards each other. On Earth, gravity gives weight to physical objects and causes them to fall towards the ground when dropped. The Earth's gravitational force is quite strong, pulling everything towards its center. The acceleration due to Earth's gravity is denoted by this constant: \( g_e = 9.81 \, \text{m/s}^2 \).

On the Moon, however, this force is significantly weaker - precisely one-sixth of Earth's gravity. This means that objects on the Moon, including astronauts jumping, experience much less gravitational pull. As a result, any outward force, such as a jump, lasts longer and covers more distance. Hence, you could jump much farther on the Moon.

Understanding gravity is crucial for calculating projectile motion, which includes any motion involving a launched object moving under the influence of gravity alone.
Range Formula
The range formula in projectile motion is used to calculate how far an object will travel horizontally when launched. It is crucial when comparing movements under different gravitational conditions.

The formula is:
  • \( R = \frac{v^2 \sin(2\theta)}{g} \)
Where:
  • \( R \) is the range of the jump.
  • \( v \) is the initial velocity or speed at takeoff.
  • \( \theta \) is the angle at which the object is launched.
  • \( g \) is the acceleration due to gravity.
By using this formula, we can calculate and compare the ranges of jumps on differing worlds or under varying gravitational pulls. The key takeaway is that a lower gravitational force increases the range of a projectile.

This formula assumes that air resistance is negligible and that the launch and landing heights are the same.
Acceleration Due to Gravity
The term "acceleration due to gravity" specifies the rate at which an object increases its velocity as it falls freely towards a celestial body. On Earth, this rate is about \( 9.81 \, \text{m/s}^2 \), indicating that an object's speed increases by 9.81 meters per second every second it is falling.

On the Moon, things are different. Here, the acceleration due to gravity is only one-sixth of Earth's, approximately \( 1.63 \, \text{m/s}^2 \). This means an object on the Moon will accelerate less quickly, reaching slower speeds over the same time span compared to the Earth.

Understanding this difference helps explain why a simple leap can put an astronaut so far ahead on lunar soil. The reduced gravitational acceleration allows a person to sustain the upward and forward momentum much longer than the same jump would on Earth. This slower "pull" is what enables greater jumps.

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Most popular questions from this chapter

When Babe Ruth hit a homer over the 8.0 -m-high rightfield fence \(98 \mathrm{~m}\) from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit \(1.0 \mathrm{~m}\) above the ground and its path initially made a \(36^{\circ}\) angle with the ground.

(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.

An object, which is at the origin at time \(t=0\), has initial velocity \(\overrightarrow{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) and constant acceleration \(\overrightarrow{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Find the position \(\overrightarrow{\mathbf{r}}\) where the object comes to rest (momentarily).

(II) An object, which is at the origin at time \(t=0,\) has $$ \overline{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} $$ $$ \quad \overline{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} . \text { Find the position } \overline{\mathbf{r}} $$ where the object comes to rest (momentarily).

\((a)\) A long jumper leaves the ground at \(45^{\circ}\) above the horizontal and lands \(8.0 \mathrm{~m}\) away. What is her "takeoff' speed \(v_{0} ?(b)\) 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. \(3-43\) )?

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