Problem 51
Question
A professional golfer can hit a ball with a speed of 70.0 \(\mathrm{m} / \mathrm{s}\) . What is the maximum distance a golf ball hit with this speed could travel on Mars, where the value of \(g\) is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) ? (The distances golf balls travel on earth are greatly shortened by air resistance and spin, as well as by the stronger force of gravity.)
Step-by-Step Solution
Verified Answer
Approximately 1320.75 meters.
1Step 1: Understand the Problem
To solve for the maximum distance a golf ball can travel, we analyze the motion as a projectile. The initial speed is given as 70.0 m/s and the acceleration due to gravity on Mars is 3.71 m/s². We need to find the maximum horizontal distance for given initial conditions.
2Step 2: Apply Projectile Motion Formula
When launched at an angle of 45° (optimal angle for maximum distance), projectile motion is symmetrical and happens as highest range. Using the formula for range \[R = \frac{v^2 \sin(2\theta)}{g}\], where \(v = 70.0 \, \text{m/s}, \theta = 45°\), \(g = 3.71 \, \text{m/s}^2\).
3Step 3: Calculate the Sine of Double the Angle
The angle \(\theta\) is 45°, hence \[\sin(2\theta) = \sin(90°) = 1\].
4Step 4: Solve for Maximum Distance on Mars
Substitute the values into the range formula:\[R = \frac{(70.0)^2 \times 1}{3.71}\]. Simplifying, calculate:\[R = \frac{4900}{3.71} \approx 1320.75 \, \text{m}\].
5Step 5: Interpret the Result
The calculated range of approximately 1320.75 meters is the maximum distance a golf ball can travel on Mars given the conditions.
Key Concepts
Mars gravityOptimal launch angleRange formulaPhysics problem solving
Mars gravity
Gravity on Mars is considerably weaker compared to Earth. While Earth's gravity is approximately 9.81 m/s², Mars exhibits a gravitational acceleration of merely 3.71 m/s². This implies that objects in free-fall on Mars experience a much lesser force pulling them down.
For projectiles, this reduced gravity means they can travel farther horizontally given the same initial speed. This is because the downward pull is weaker, allowing the projectile more time in the air. As a result, the maximum distance a projectile can travel becomes significantly greater.
Understanding Mars gravity is essential when solving projectile problems as it directly influences both the motion path and the range of the projectile.
For projectiles, this reduced gravity means they can travel farther horizontally given the same initial speed. This is because the downward pull is weaker, allowing the projectile more time in the air. As a result, the maximum distance a projectile can travel becomes significantly greater.
Understanding Mars gravity is essential when solving projectile problems as it directly influences both the motion path and the range of the projectile.
Optimal launch angle
One of the crucial factors in maximizing the range of a projectile is the launch angle. Intuitively, we might think that a steeper angle would launch a projectile further, but there is a specific angle that achieves the maximum distance.
In the absence of air resistance, the optimal launch angle for maximum range is 45 degrees. At this angle, the projectile's motion is effectively balanced between vertical height and horizontal distance. The launch angle influences the time the projectile remains in the air and the initial speed distribution in vertical and horizontal components.
On planets like Mars, the optimal angle remains 45 degrees because the force of gravity still acts downwards uniformly, and air resistance is negligible in this consideration.
In the absence of air resistance, the optimal launch angle for maximum range is 45 degrees. At this angle, the projectile's motion is effectively balanced between vertical height and horizontal distance. The launch angle influences the time the projectile remains in the air and the initial speed distribution in vertical and horizontal components.
On planets like Mars, the optimal angle remains 45 degrees because the force of gravity still acts downwards uniformly, and air resistance is negligible in this consideration.
Range formula
The range formula is vital for determining how far a projectile will travel when launched. The formula is given by:\[ R = \frac{v^2 \sin(2\theta)}{g} \]where:
Inserting the recognized optimal launch angle of 45 degrees simplifies the formula since \( \sin(90°) = 1 \). This is what makes calculations for maximum range simpler, particularly in an environment like Mars where gravity is reduced.
- \( R \) is the range or maximum horizontal distance
- \( v \) represents initial velocity
- \( \theta \) is the launch angle
- \( g \) is the acceleration due to gravity
Inserting the recognized optimal launch angle of 45 degrees simplifies the formula since \( \sin(90°) = 1 \). This is what makes calculations for maximum range simpler, particularly in an environment like Mars where gravity is reduced.
Physics problem solving
Solving physics problems, particularly involving projectile motion, requires a comprehensive understanding of underlying principles and systematic approaches. Here's a step-by-step guide to solving such problems:
- **Understand the problem:** Analyze what is given and what is required. Ensure all units are consistent.
- **Identify key components:** Recognize initial velocity, angle of launch, and gravitational force.
- **Apply suitable formulas:** Depending on what needs to be solved (e.g. distance, time, height), choose the relevant equations.
- **Make calculations:** Insert known values into formulas and compute results.
- **Interpret results:** Make sense of your findings in the context of the problem.
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