Problem 112 - Fundamentals of Physics

Question

The range of a projectile depends not only on \(v_{0}\) and \(\theta_{0}\) but also on the value \(g\) of the free-fall acceleration, which varies from place to place. In 1936 , Jesse Owens established a world's running broad jump record of \(8.09 \mathrm{~m}\) at the Olympic Games at Berlin (where \(g=9.8128 \mathrm{~m} / \mathrm{s}^{2}\) ). Assuming the same values of \(v_{0}\) and \(\theta_{0}\), by how much would his record have differed if he had competed instead in 1956 at Melbourne (where \(\left.g=9.7999 \mathrm{~m} / \mathrm{s}^{2}\right)\) ?

Step-by-Step Solution

Answer

The Melbourne jump would be approximately 1.4 mm longer than in Berlin.

Step 1: Understand the Physics

The range of a projectile launched at an initial speed \( v_0 \) and angle \( \theta_0 \) is given by \[R = \frac{v_0^2 \sin(2\theta_0)}{g},\]where \( g \) is the acceleration due to gravity.

Step 2: Identify Known Values in Berlin

For the Berlin Olympics, Jesse Owens' jump was \[ R_{Berlin} = 8.09 \, \text{m} \]and the gravitational acceleration was \[ g_{Berlin} = 9.8128 \, m/s^2.\]

Step 3: Express Initial Velocity and Angle

Using the formula for range and the known values from Berlin:\[ 8.09 = \frac{v_0^2 \sin(2\theta_0)}{9.8128}. \]Rearrange to express \[ v_0^2 \sin(2\theta_0) = 8.09 \times 9.8128. \]

Step 4: Calculate the Product of Initial Velocity and Angle

Calculate the product \[ v_0^2 \sin(2\theta_0) \approx 79.290592. \]

Step 5: Calculate Range in Melbourne

For Melbourne, use the same initial conditions but different gravity:\[ R_{Melbourne} = \frac{v_0^2 \sin(2\theta_0)}{9.7999} \approx \frac{79.290592}{9.7999}. \]

Step 6: Compute Final Result

Calculate \[ R_{Melbourne} \approx 8.0914 \, \text{m}. \]The difference from Berlin's range is:\[ R_{Melbourne} - R_{Berlin} \approx 8.0914 - 8.09 \approx 0.0014 \, \text{m} \]or 1.4 mm.

Key Concepts

Free-fall Acceleration

Free-fall acceleration, symbolized as \( g \), represents the acceleration of an object caused by the force of gravity. It is crucial in projectile motion as it influences how quickly an object accelerates downward. On Earth, \( g \) typically measures approximately \( 9.81 \, \text{m/s}^2 \). However, slight variations occur depending on geographic location due to differences in altitude, Earth's rotation, and local geology.
Understanding these small differences is important as they can affect the range of a projectile. In our exercise, the change from Berlin's \( g = 9.8128 \, \text{m/s}^2 \) to Melbourne's \( g = 9.7999 \, \text{m/s}^2 \) had a subtle impact on Jesse Owens' jump.

• These variations can solve why two jumps with the same initial conditions can have slightly different distances.
• In projectile motion, a smaller \( g \) results in longer hang time, slightly increasing the range.

Therefore, acknowledging the value of \( g \) is vital for precise calculations in physics problems.

Range of a Projectile

The range of a projectile refers to the horizontal distance it travels during its motion. For a given initial speed and launch angle, the range \( R \) can be determined using the equation:\[ R = \frac{v_0^2 \sin(2\theta_0)}{g} \]In this equation:

• \( v_0 \) is the initial velocity.
• \( \theta_0 \) is the launch angle.
• \( g \) is the free-fall acceleration.

Notice how \( g \) is in the denominator, meaning changes in gravity directly influence the range. A stronger gravitational pull compresses the range, while a weaker pull extends it.
Our exercise highlighted this concept by comparing projectile ranges at different gravitation values, leading to a small but noticeable deviation in Jesse Owens' jump.
To optimize range, it is crucial to consider factors like launch angle and initial speed while being mindful of local variations in \( g \).

Initial Velocity and Launch Angle

When discussing projectile motion, initial velocity \( v_0 \) and launch angle \( \theta_0 \) are essential. They determine the projectile's trajectory, speed, and ultimately the range. Initial velocity refers to the speed of the object at the moment of release, while the launch angle is the angle at which the object is propelled relative to the horizontal.
Calculations in our exercise show how these two factors combine to yield the product \( v_0^2 \sin(2\theta_0) \), an essential component of the range equation.

• Higher initial velocity increases the potential range.
• The ideal launch angle for maximum range, in the absence of air resistance, is \( 45^\circ \).

When tackling problems, it can be useful to:

• Visualize the trajectory, which forms a parabolic path.
• Recognize optimal conditions for maximizing distance.

This approach helps break down the problem and find accurate solutions. Thus, analyzing both the initial velocity and launch angles provides a deeper understanding of projectile motion.