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Chapter 3: Problem 19

(II) Estimate by what factor a person can jump farther 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 6 times farther on the Moon than on Earth.

Step by step solution

01

Understand the Problem

We are given two scenarios: jumping on Earth and jumping on the Moon. We need to compare how the distances differ given the same takeoff speed and angle. The acceleration due to gravity is different in these two cases, which affects the jumping distance.
02

Recall the Formula for Range

The horizontal range of a projectile, such as a jump, 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

Compare Gravity on Earth and Moon

Let \( g_e \) denote the acceleration due to gravity on Earth, and \( g_m = \frac{1}{6}g_e \) denote that on the Moon. This relationship is crucial because the range is inversely proportional to the acceleration due to gravity.
04

Relate Range on Moon to Range on Earth

Using the range formula, the range on Earth \( R_e \) and the range on the Moon \( R_m \) are given by:\[ R_e = \frac{v^2 \sin(2\theta)}{g_e} \]\[ R_m = \frac{v^2 \sin(2\theta)}{g_m} \]Substitute \( g_m = \frac{1}{6}g_e \) into the second equation:\[ R_m = \frac{v^2 \sin(2\theta)}{\frac{1}{6}g_e} = 6 \times \frac{v^2 \sin(2\theta)}{g_e} \]
05

Calculate the Factor Increase

From the expressions, we see that:\[ R_m = 6 \times R_e \]This indicates that the range on the Moon is six times the range on Earth, given the same conditions.

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Key Concepts

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

Acceleration Due to Gravity
Gravity is the force that pulls objects towards the center of a celestial body, like the Earth or the Moon. It plays a crucial role in projectile motion by affecting the path and distance an object travels. - On Earth, the acceleration due to gravity \( g \) is approximately \( 9.81 \text{ m/s}^2 \).- This means any object in free fall will accelerate towards the Earth at this rate.- On the Moon, gravity is significantly weaker due to its smaller mass and size. This makes \( g \) on the Moon about one-sixth of that on Earth, or approximately \( 1.63 \text{ m/s}^2 \).The reduced gravitational pull on the Moon means that projectiles, such as a jump, experience less downward force, enabling them to travel further before hitting the ground.
Range Formula
The range formula is essential in predicting how far a projectile will travel. When analyzing jumps, it helps to calculate the distance covered when an object launches at a certain angle and speed.The formula for the horizontal range \( R \) of a projectile is:\[ R = \frac{v^2 \sin(2\theta)}{g} \]where:- \( v \) is the initial velocity,- \( \theta \) is the launch angle,- \( g \) is the acceleration due to gravity. This equation shows that:- The range is directly proportional to the square of the initial speed and the sine of double the launch angle.- It is inversely proportional to the acceleration due to gravity. So, a smaller \( g \) results in a larger range.This concept helps us understand why a jump on the Moon can cover more ground compared to the same jump on Earth.
Moon Versus Earth Comparison
Comparing jumps on the Moon and Earth reveals fascinating insights into Earth's stronger gravitational force. If a person jumps with the same speed and angle on both celestial bodies, the range differs due to the disparity in gravitational pull.- On Earth, the stronger gravity \( g_e \) limits the distance a jumper can achieve.- On the Moon, with a gravity \( g_m \) being \( \frac{1}{6}g_e \), the range increase is significant.By substituting the ratio \( g_m = \frac{1}{6}g_e \) into the range formula for the Moon, we find:\[ R_m = 6 \times R_e \]This shows a person can jump six times farther on the Moon than on Earth under identical conditions. The weaker lunar gravity allows for a longer airborne time, translating to a much greater horizontal distance.

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Most popular questions from this chapter

Spymaster Chris, flying a constant 208 km/h horizontally in a low-flying helicopter, wants to drop secret documents into her contact's open car which is traveling 156 km/h on a level highway 78.0 m below. At what angle (with the horizontal) should the car be in her sights when the packet is released (Fig. 3-55)?

(II) \(\overrightarrow {V}\) is a vector 24.8 units in magnitude and points at an angle of 23.4\(^\circ\) above the negative \(x\) axis. (\(a\)) Sketch this vector. (\(b\)) Calculate \(V_x\) and \(V_y\) (\(c\)) Use \(V_x\) and \(V_y\) to obtain (again) the magnitude and direction of \(\overrightarrow {V}\). [\(Note\): Part (\(c\)) is a good way to check if you've resolved your vector correctly.]

(II) Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 4.0 m/s and enjoys a free fall until she is 150 m above the valley floor, at which time she opens her parachute (Fig. 3-37). (\(a\)) How long is the jumper in free fall? Ignore air resistance. (\(b\)) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?

A boat can travel 2.20 m/s in still water. (\(a\)) If the boat points directly across a stream whose current is 1.20 m/s, what is the velocity (magnitude and direction) of the boat relative to the shore? (\(b\)) What will be the position of the boat, relative to its point of origin, after 3.00 s?

If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what % will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18m away? For this estimate, ignore air resistance and spin on the ball.
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