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(a) If a person can jump a maximum horizontal distance (by using a \(45^{\circ}\) projection angle) of \(3.0 \mathrm{~m}\) on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is \(g / 6\) and \(g=\) \(9.80 \mathrm{~m} / \mathrm{s}^{2}\) ? (b) Repeat for Mars, where the acceleration due to gravity is \(0.38 \mathrm{~g}\).

Short Answer

Expert verified
The maximum jump range on the Moon is 18m, and on Mars it is approximately 7.9m.

Step by step solution

01

Understanding the problem

We know the maximum range on Earth and we know the different strengths of gravity on the Moon and Mars compared to Earth. We use the projectile motion formula for the range: \(R = \frac{v^2}{g} \sin(2\Theta)\), where \(R\) is the range, \(v\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(\Theta\) is the angle of projection. Since the angle and initial speed don't change, we just need to adjust for the altered gravity when calculating the range on the Moon and Mars.
02

Calculate Range on the Moon

The gravity on the Moon is \(g / 6\), so the modified equation for the horizontal distance on the Moon is: \(R_{moon} = 6R_{earth} = 6 * 3m = 18m\).
03

Calculate Range on Mars

Similarly, for Mars, the gravity is \(0.38 \times g\), i.e., 38% of Earth's gravity. We substitute into the equation for the range: \(R_{mars} = \frac{1}{0.38}*R_{earth} = \frac{1}{0.38} * 3m \approx 7.9m\).

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Key Concepts

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

Free-fall Acceleration
Free-fall acceleration is a key concept in understanding how objects move under the influence of gravity alone. On Earth, the standard free-fall acceleration is denoted by the symbol \( g \) and has a value of approximately \( 9.80 \text{ m/s}^2 \). This value indicates the rate of change of velocity of an object when it is allowed to fall freely in a vacuum, without air resistance.
In simpler terms, if you drop a ball from a height, it will accelerate towards the ground at \( 9.80 \text{ m/s}^2 \) until it impacts the ground. Free-fall acceleration is crucial for calculating various motions and behaviors of projectiles, such as their speed, time of flight, and range.
Gravity Comparison
Gravity differs significantly across celestial bodies. It is an essential factor when calculating projectile motion since it directly influences how far an object can travel. The exercise involves comparing Earth's gravity with that of the Moon and Mars.
  • Moon: The free-fall acceleration on the Moon is only \( g/6 \), which translates to about \( 1.63 \text{ m/s}^2 \). This is why astronauts are able to jump much higher on the Moon compared to Earth.
  • Mars: Mars has a gravity that is \( 0.38 \) times that of Earth. This means its free-fall acceleration is approximately \( 3.7 \text{ m/s}^2 \).
Understanding these differences is vital for making accurate calculations regarding the range and flight dynamics of projectiles on these celestial bodies.
Range Equation
The range of a projectile, which is the horizontal distance it travels, depends on several factors including initial velocity, launch angle, and gravitational acceleration. The equation used for calculating this is formulated as:\[R = \frac{v^2}{g} \sin(2\Theta)\]where:
  • \( R \) is the range.
  • \( v \) is the initial velocity of the projectile.
  • \( g \) is the acceleration due to gravity.
  • \( \Theta \) is the angle of projection.
The \( \sin(2\Theta) \) component reaches its maximum value of 1 when \( \Theta = 45^{\circ} \), which is why a \( 45^{\circ} \) angle is used for maximum range calculations. Knowing how to adjust for different gravitational environments, such as those on the Moon or Mars, is essential for engineering and scientific applications.
Planetary Gravity
Planetary gravity refers to the gravitational force exerted by a planet or celestial body. This gravitational pull varies from one body to another and directly affects projectile motion by altering the acceleration factor in the range equation. Different planetary bodies have unique gravitational forces, leading to distinct environmental conditions.
Understanding planetary gravity helps predict how objects behave in space exploration missions. For example, if an astronaut were to launch a projectile on different planets, they would experience different ranges due to the unique free-fall accelerations:
  • Earth: \( 9.80 \text{ m/s}^2 \)
  • Moon: \( 1.63 \text{ m/s}^2 \)
  • Mars: \( 3.7 \text{ m/s}^2 \)
Designing equipment and planning missions requires accounting for these differences to ensure successful outcomes.
Projectile Range Calculation
Projectile range calculation is an application of physics principles that involves determining how far a projectile can travel. To calculate projectile range, one uses the range equation, which is affected by the initial velocity, projection angle, and planet's gravity.
In our exercise example:
  • On the Moon: With the weaker gravity of \( g/6 \), the calculated range increases significantly to 18 meters when the same jump is performed on the Moon. This is because the reduced gravitational force allows the projectile to travel further.
  • On Mars: With gravity being \( 0.38 \) times Earth's, the range calculation yields about 7.9 meters. Mars' gravity, being less than Earth's but more than the Moon's, results in a moderate increase in range.
Calculations like these demonstrate the adaptability needed for missions on different celestial bodies, impacting everything from sports played to mechanisms for moving across planetary surfaces.

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Most popular questions from this chapter

A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range \(R\). (a) At what angle \(\theta\) is the rock thrown? (b) In terms of the original range \(R\), what is the range \(R_{\max }\) the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.

A playground is on the flat roof of a city school, \(6.00 \mathrm{~m}\) above the street below (Fig. P3.34). The vertical wall of the building is \(h=7.00 \mathrm{~m}\) high, to form a l-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of \(\theta=53.0^{\circ}\) above the horizontal at a point \(d=24.0 \mathrm{~m}\) from the base of the building wall. The ball takes \(2.20\) s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

The record distance in the sport of throwing cowpats is \(81.1 \mathrm{~m}\). This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was \(45^{\circ}\) and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time the projectile was in flight. (c) Qualitatively, how would the answers change if the launch angle were greater than \(45^{\text {" }}\) Explain.

A bolt drops from the ceiling of a moving train car that is accelerating northward at a rate of \(2.50 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the acceleration of the bolt relative to the train car? (b) What is the acceleration of the bolt relative to the Earth? (c) Describe the trajectory of the bolt as seen by an observer fixed on the Earth.

A home run is hit in such a way that the baseball just clears a wall \(21 \mathrm{~m}\) high, located \(130 \mathrm{~m}\) from home plate. The ball is hit at an angle of \(35^{\circ}\) to the horizontal, and air resistance is negligible. Find (a) the initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of \(1.0 \mathrm{~m}\) above the ground.)

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