Problem 22 - Thomas Calculus

Question

Throwing a baseball A baseball is thrown from the stands 32 \(\mathrm{ft}\) above the field at an angle of \(30^{\circ}\) up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 32 \(\mathrm{ft} / \mathrm{sec}\) ?

Step-by-Step Solution

Answer

The ball hits the ground after 2 seconds, about 55.42 feet away.

Step 1: Identify the Problem Variables

We need to calculate the time (t) it takes for a baseball projected from a height of 32 ft to hit the ground and the horizontal distance (range) it covers. Given: initial height = 32 ft, initial speed = 32 ft/sec, and launch angle = 30°.

Step 2: Resolve Initial Speed into Components

Calculate the horizontal (\(v_{0x}\)) and vertical components (\(v_{0y}\)) of the initial velocity:\[v_{0x} = v_0 \cos(30^\circ) = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3} \approx 27.71 \, \text{ft/sec}\]\[v_{0y} = v_0 \sin(30^\circ) = 32 \times \frac{1}{2} = 16 \, \text{ft/sec}\]

Step 3: Calculate Time to Hit the Ground

Use the vertical motion equation considering downward acceleration due to gravity:\[y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2\]Set \(y(t) = 0\) (ground level), where \(g = 32\, \text{ft/sec}^2\):\[0 = 32 + 16t - 16t^2\]Solve the quadratic equation:\[-16t^2 + 16t + 32 = 0\]Factor or use the quadratic formula to find \(t\); the positive root is relevant:\[t = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} = \frac{-16 ± \sqrt{16^2 + 4\times 16\times 16}}{2(-16)}\]Simplify to find \(t\): approximately \(2\) seconds.

Step 4: Calculate Horizontal Range

Calculate the horizontal displacement (range) using the horizontal velocity component and time:\[\text{Range} = v_{0x} \cdot t = 16\sqrt{3} \times 2\]the range is approximately \(55.42\, \text{ft}\).

Step 5: Verify Calculations and Conclusion

Double-check the components and calculations ensuring units and operations, especially signs in the quadratic, are correctly handled. Ensure the use of the only physically meaningful solution for time.

Key Concepts

Quadratic Equation

In projectile motion problems, like throwing a baseball, a quadratic equation often emerges when resolving the vertical motion of the projectile. This is because the vertical movement is influenced by gravity, creating a parabolic path.

The general form of a quadratic equation is expressed as:

• \( ax^2 + bx + c = 0 \)

Here, the coefficients \( a \), \( b \), and \( c \) represent constants, while \( x \) is the variable representing time or another parameter. In our specific problem, the vertical motion is described by:

• \( -16t^2 + 16t + 32 = 0 \)
• Where \( t \) is the time in seconds, and \( g = 32 \text{ ft/s}^2 \) is the acceleration due to gravity.

To solve for \( t \), use the quadratic formula:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• The positive root of this equation helps determine how long the baseball stays in the air.

Understanding how to handle the quadratic in these problems is crucial for predicting when a projectile reaches the ground.

Velocity Components

When dealing with projectile motion, understanding velocity components is critical. These components break down the initial velocity into horizontal and vertical directions, which behave differently in motion.

The initial speed in projectile motion can be divided using trigonometry:

• Horizontal component \( v_{0x} \)
• Vertical component \( v_{0y} \)

The formulas used are:

• \( v_{0x} = v_0 \cos(\theta) \)
• \( v_{0y} = v_0 \sin(\theta) \)

Where \( v_0 \) is the initial velocity and \( \theta \) is the angle of projection. In our problem, the baseball's initial speed of 32 ft/sec at a 30-degree angle results in:

• \( v_{0x} = 27.71 \text{ ft/sec} \)
• \( v_{0y} = 16 \text{ ft/sec} \)

These components are essential for calculating both how far and when the projectile lands.

Gravity

Gravity is the force that pulls the baseball back towards the Earth, creating a parabolic trajectory, typical of projectile motion. This force acts vertically downward at \( g = 32\, \text{ft/sec}^2 \) in our exercise context.

Gravity impacts the vertical motion of the baseball:

• The downward acceleration influences how long the baseball stays airborne.
• It decreases the upward vertical velocity after the ball is thrown until reaching its peak.
• Once the peak is reached, the ball descends, accelerating due to gravity until it hits the ground.

It is gravity that necessitates the use of the quadratic equation to analyze the baseball's movement. By understanding how gravity interacts with velocity components, we can accurately predict the projectile's landing time and distance.