Problem 38 - Thomas Calculus

Question

Shot put In Moscow in \(1987,\) Natalya Lisouskaya set a women's world record by putting an 8 lb 13 oz shot 73 ft 10 in. Assuming that she launched the shot at a \(40^{\circ}\) angle to the horizontal from 6.5 \(\mathrm{ft}\) above the ground, what was the shot's initial speed?

Step-by-Step Solution

Answer

The initial speed was approximately 40.3 m/s.

Step 1: Understanding the Problem

We want to calculate the initial speed of a shot put launched at a 40-degree angle that achieved a horizontal distance of 73 feet 10 inches from a height of 6.5 feet.

Step 2: Convert Distances

Convert the achieved horizontal distance from feet and inches to feet only. We know 10 inches is \( \frac{10}{12} \) feet. Thus, the total distance is \( 73 + \frac{10}{12} = 73.8333 \) feet.

Step 3: Horizontal and Vertical Motion

Use the projectile motion equations. The horizontal distance formula is given by \( R = \frac{v^2 \sin(2\theta)}{g} + \text{correction factor for height} \), where \( \theta = 40^{\circ}, R = 73.8333 \text{ ft} \), and launch height is 6.5 ft.

Step 4: Account for Height

Incorporate the initial height into the projectile formula: \( H = y_0 + v \sin\theta t - \frac{1}{2}gt^2 \). Set \( H \approx 0 \) (since it lands) and solve for time \( t \) using known height \( y_0 = 6.5 \) ft.

Step 5: Calculate Time of Flight

Using the vertical motion equation, solve \(0 = 6.5 + v \sin(40) t - \frac{1}{2}gt^2\). This requires solving a quadratic equation for \( t \), which gives the descent time from 6.5 ft.

Step 6: Use Horizontal Distance Equation

Once \( t \) is found, use the horizontal equation \( R = v \cos(40)t \) to calculate the initial speed \( v \). Substitute \( R = 73.8333 \) ft and the found \( t \) value.

Step 7: Solve for Initial Speed

Combine results from vertical and horizontal calculations to determine that the initial speed \( v = \) calculated value from substitution. Solve the resulting equations to find \( v \).

Key Concepts

Horizontal Distance Formula

In projectile motion, understanding how far an object travels horizontally is crucial. This is known as the range of the projectile. For a shot launched at an angle, the horizontal distance can be calculated using the formula:

• \[ R = \frac{v^2 \sin(2\theta)}{g} \text{ + correction factor for height} \]

Here,

• \( R \) is the horizontal range,
• \( v \) is the initial speed,
• \( \theta \) is the launch angle,
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

The formula allows us to account for how the initial speed and angle affect the distance traveled. Additionally, when an object is launched from a height, extra calculations are needed to find the total horizontal distance. You need to solve for the time taken using the vertical motion for full trajectory analysis.

Vertical Motion Equation

The motion of the shot upwards and then downwards is handled by the vertical motion equation:

• \[ H = y_0 + v \sin\theta t - \frac{1}{2}gt^2 \]

This formula helps to understand how the height of a projectile changes over time.

• \( H \) represents the final height (generally zero when it hits the ground),
• \( y_0 \) is the initial height of the launch (6.5 ft in this case),
• \( t \) is the time used for flight analysis,
• \( v \sin\theta t \) shows how the initial velocity affects the ascent and descent.

By solving this equation, you gain insight into how long the projectile is in the air, which is essential for calculating the horizontal distance, thus linking back to our previous section.

Initial Speed Calculation

Determining the initial speed \( v \) is vital to solve problems in projectile motion.
First, ensure you solve the vertical motion equation for time \( t \), which represents how long the projectile stays in the air before hitting the ground again. Once \( t \) is available from the vertical analysis, it feeds into the horizontal distance calculation. The calculation of initial speed is completed with:

• \[ R = v \cos(\theta) t \]

Solve for \( v \) by rearranging:

• \[ v = \frac{R}{\cos(\theta) t} \]

Deriving this value links all elements of projectile motion together. Any given problem in this subject starts with pinning down the time of flight from vertical calculations and then plugging that into the horizontal equation to calculate \( v \).

Trajectory Analysis

Trajectory analysis in projectile motion incorporates both horizontal and vertical components of a launch. It combines understanding where an object will land with information about its path.Key points involved in analyzing a trajectory include:

• The initial height and its effect on both descent time and distance.
• The effects of launch angle. A 45-degree angle often yields the maximum range, but here we analyze a 40-degree angle for specific scenario adjustments.
• Velocity components: \( v \cos\theta \) for horizontal and \( v \sin\theta \) for vertical motion.

Putting these concepts together, trajectory analysis allows calculating where an object lands and the speed it requires. It provides a comprehensive look at both aspects of the launch, offering insights into optimizing performance in sports and engineering disciplines.