Problem 27 - Thomas Calculus

Question

Firing golf balls A spring gun at ground level fires a golf ball at an angle of \(45^{\circ} .\) The ball lands 10 \(\mathrm{m}\) away. a. What was the ball's initial speed? b. For the same initial speed, find the two firing angles that make the range 6 \(\mathrm{m} .\)

Step-by-Step Solution

Answer

a) The initial speed is 9.9 m/s. b) The angles are 18.3° and 71.7°.

Step 1: Identify the Initial Problem

First, identify the given information and what we need to find. The problem states a spring gun fires a golf ball at a 45-degree angle, landing 10 meters away. We need to determine the initial speed of the ball.

Step 2: Use the Range Formula

The range formula for projectile motion is given by \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( R \) is the range (10 m), \( v_0 \) is the initial speed, \( \theta \) is the launch angle (45 degrees), and \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)).

Step 3: Simplify the Formula for 45 Degrees

For a launch angle of 45 degrees, \( \sin(90^{\circ}) = 1 \) because \( \sin(2\times45^{\circ}) = \sin(90^{\circ}) \). This simplifies the range formula to \( R = \frac{v_0^2}{g} \).

Step 4: Solve for Initial Speed

Rearrange the formula \( R = \frac{v_0^2}{g} \) to find the initial speed \( v_0 \): \( v_0 = \sqrt{R \times g} \). Substitute the values \( R = 10 \) m and \( g = 9.8 \) m/s\(^2\) to get \( v_0 = \sqrt{10 \times 9.8} = \sqrt{98} = 9.9 \) m/s.

Step 5: Set Up Equation for New Range

For a new range of 6 m with the same initial speed, use the formula \( 6 = \frac{(9.9)^2 \sin(2\theta)}{9.8} \) and solve for \( \theta \).

Step 6: Rearrange and Solve for Angles

Rearrange to find \( \sin(2\theta) = \frac{6 \times 9.8}{(9.9)^2} = 0.5958 \). Find the angle using \( 2\theta = \sin^{-1}(0.5958) \). Calculate \( 2\theta \) to get two possible angles: \( 2\theta = 36.6^{\circ} \) and \( 2\theta = 143.4^{\circ} \) (since \( \sin \) function is positive in the first and second quadrants).

Step 7: Determine Final Angles

Divide \( 2\theta \) by 2 to get the launch angles: \( \theta = 18.3^{\circ} \) and \( \theta = 71.7^{\circ} \).

Key Concepts

Launch Angle

The launch angle, denoted as \( \theta \), in projectile motion is the angle at which an object, such as a golf ball, is fired or launched into the air. This angle is crucial because:

• It influences the range, height, and time of flight of the projectile.
• Maximizing the range for a given speed typically occurs at a 45-degree angle.
• Changes in the launch angle, while keeping other variables the same, affect where the projectile will land.

For example, in the exercise, the golf ball is initially fired at a 45-degree angle, which is ideal for achieving maximum distance. Furthermore, to achieve different target distances (like 6 meters), finding the correct launch angles (here, \(18.3^{\circ}\) and \(71.7^{\circ}\)) is necessary. These angles demonstrate that for a fixed initial speed, there can be two different launch angles that allow the projectile to reach the same point - a lower angle with less height and a higher angle with more height.

Range Formula

The range formula is a cornerstone of projectile motion, connecting various factors like speed, angle, and gravity:\[ R = \frac{v_0^2 \sin(2\theta)}{g} \] where:

• \( R \) is the range, or horizontal distance traveled.
• \( v_0 \) is the initial speed of the projectile.
• \( \theta \) is the launch angle.
• \( g \) is the acceleration due to gravity.

For a launch angle of 45 degrees, the sine function reaches its maximum value, \( \sin(90^{\circ}) = 1 \), simplifying the formula to \( R = \frac{v_0^2}{g} \). This formula is used to solve for various unknowns. In our example, it was crucial to determine the initial speed needed to achieve a 10-meter range and to adjust launch angles for a different 6-meter target.

Initial Speed

Initial speed, denoted as \( v_0 \), is the speed at which a projectile is launched. It's a vital component in determining how far or high the projectile will travel. Using the range formula, we can derive the initial speed needed to reach a specific range given other parameters. In the equation rearrangement: \[ v_0 = \sqrt{R \times g} \]Using this formula, and knowing the range and acceleration due to gravity, you can calculate the initial speed. For example, to land a golf ball 10 meters away with a launch angle of 45 degrees:

• \( R = 10 \) meters
• \( g = 9.8 \) m/s²

The initial speed comes out to be approximately 9.9 m/s.

Acceleration due to Gravity

Acceleration due to gravity, commonly denoted as \( g \), is a constant that measures the rate of attraction between objects, primarily felt as the force pulling objects back to Earth. It is approximately \( 9.8 \) m/s\(^2\) on Earth and plays a key role in:

• The time the projectile stays in the air (time of flight).
• The maximum height reached by the projectile.
• Overall distance or range traveled by the projectile.

In our example of projectile motion, knowing \( g \) allows us to accurately apply the range formula to calculate initial speed and solve for different launch angles. It is crucial for achieving precise calculations in physics, ensuring the golf ball lands exactly where intended.