Problem 40 - Fundamentals of Physics

Question

Suppose that a shot putter can put a shot at the worldclass speed \(v_{0}=15.00 \mathrm{~m} / \mathrm{s}\) and at a height of \(2.160 \mathrm{~m}\). What horizontal distance would the shot travel if the launch angle \(\theta_{0}\) is (a) \(45.00^{\circ}\) and (b) \(42.00^{\circ}\) ? The answers indicate that the angle of \(45^{\circ}\), which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Step-by-Step Solution

Answer

Range is longest at 42° with approximately 22.78 m.

Step 1: Understanding the Formula

To calculate the horizontal distance, we need to use the projectile motion equation for range when the initial and final heights are different:\[R = \frac{v_0\cos(\theta_0)}{g}\left(v_0\sin(\theta_0) + \sqrt{(v_0\sin(\theta_0))^2 + 2g h}\right)\]where \( R \) is the range, \( v_{0} \) is the initial velocity, \( \theta_{0} \) is the launch angle, \( g \) is the acceleration due to gravity \((9.81 \text{ m/s}^2)\), and \( h \) is the initial height.

Step 2: Calculate Range for 45°

Now we're going to substitute the given values into the formula for a launch angle \( \theta_{0} = 45^{\circ} \):\[R = \frac{15.00\cos(45^{\circ})}{9.81}\left(15.00\sin(45^{\circ}) + \sqrt{(15.00\sin(45^{\circ}))^2 + 2 \cdot 9.81 \cdot 2.160}\right)\]Calculating each part:- \( \cos(45^{\circ}) = \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \)Plug these into the equation, simplifying:\[R \approx \frac{15.00\times\frac{\sqrt{2}}{2}}{9.81}\left(15.00\times\frac{\sqrt{2}}{2} + \sqrt{(15.00\times\frac{\sqrt{2}}{2})^2 + 2 \cdot 9.81 \cdot 2.160}\right)\]Solving this gives the range \( R \approx 22.59 \text{ m} \).

Step 3: Calculate Range for 42°

Next, we'll calculate the range for a launch angle \( \theta_{0} = 42^{\circ} \):\[R = \frac{15.00\cos(42^{\circ})}{9.81}\left(15.00\sin(42^{\circ}) + \sqrt{(15.00\sin(42^{\circ}))^2 + 2 \cdot 9.81 \cdot 2.160}\right)\]Compute \( \cos(42^{\circ}) \approx 0.743 \) and \( \sin(42^{\circ}) \approx 0.669 \):Substitute these into the equation:\[R \approx \frac{15.00 \times 0.743}{9.81} \left(15.00 \times 0.669 + \sqrt{(15.00 \times 0.669)^2 + 2 \cdot 9.81 \cdot 2.160}\right)\]Calculate the above expression for the horizontal distance, which gives \( R \approx 22.78 \text{ m} \).

Step 4: Compare the Distances

Compare both the calculated ranges:- Range at \( 45^{\circ} \) is approximately \( 22.59 \text{ m} \).- Range at \( 42^{\circ} \) is approximately \( 22.78 \text{ m} \).This shows that the shot put travels farther with a launch angle of \( 42^{\circ} \) than with \( 45^{\circ} \), demonstrating that the classic 45° angle doesn't maximize the horizontal distance when launch and landing heights differ.

Key Concepts

Range Equation

The range equation is key to understanding projectile motion, which describes the path a projectile takes through space. In situations where the initial and final heights differ, as is the case with a shot put, the range equation helps calculate how far the projectile will travel horizontally.
The equation is given by: \[R = \frac{v_0\cos(\theta_0)}{g}\left(v_0\sin(\theta_0) + \sqrt{(v_0\sin(\theta_0))^2 + 2g h}\right)\]
In this equation:

• \( R \) denotes the range, or horizontal distance, of the projectile.
• \( v_0 \) is the initial velocity at which the projectile is launched.
• \( \theta_0 \) is the launch angle.
• \( g \) represents the acceleration due to gravity, constant at approximately \( 9.81 \text{ m/s}^2 \).
• \( h \) is the initial height from which the projectile is released.

The formula incorporates both components of velocity (horizontal and vertical). Calculating range requires substituting the specific parameters into the equation. For students, recognizing how various factors like angle and height affect range can be crucial in comprehending projectile motion.

Launch Angle

The launch angle, \( \theta_0 \), is a fundamental factor in projectile motion that determines the trajectory and range of a projectile. It is the angle at which the projectile is released relative to the horizontal.
The launch angle is significant because it influences how both the horizontal and vertical components of velocity impact motion.- At a launch angle of \( 45^{\circ} \), theoretically the projectile would travel the maximum possible horizontal distance in a world without differing launch and landing heights.- However, when initial and final heights differ, as in many real-world applications, the angle that maximizes the horizontal distance can vary.In our exercise, we discovered that a \( 42^{\circ} \) angle resulted in a greater range compared to the classic \( 45^{\circ} \), providing a practical example of how angle adjustments can better suit particular conditions.

Initial Velocity

Initial velocity, \( v_0 \), is the speed at which a projectile is launched into the air. It is an essential component of projectile motion, impacting both the range and direction of the projectile.
In our scenario:

• The initial velocity is given as \( 15.00 \text{ m/s} \).
• This velocity is decomposed into horizontal and vertical components using the launch angle and trigonometric functions \( \cos(\theta_0) \) and \( \sin(\theta_0) \).
• These components play a key role in determining the distance and motion path taken by the projectile.

Understanding how the initial velocity splits into components helps reveal why slight changes in launch angle can significantly alter the range. It emphasizes how both speed and angle affect the overall motion of a projectile.

Gravity Effect

Gravity is a constant force acting on projectiles, pulling them toward the Earth. Its effect on projectile motion is crucial as it influences the path, velocity, and range of the projectile.
Some points include:

• The acceleration due to gravity is constant at approximately \( 9.81 \text{ m/s}^2 \).
• Gravity affects the vertical component of the projectile's velocity, causing it to decelerate as it rises and accelerate as it falls.
• It also determines how long the projectile remains airborne, directly impacting the horizontal distance traveled.

In the projectile motion formula, gravity helps calculate the time of flight and trajectory, integrating with initial velocity and launch angle to establish the projectile’s path. Recognizing gravity's role allows students to predict how different launch conditions will alter the projectile's range and flight characteristics.