Problem 23 - 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 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

Initial speed is approximately 9.90 m/s. Firing angles are 18.44° and 71.56°.

Step 1: Understand the problem

We're dealing with projectile motion. A golf ball is fired at a 45° angle and lands 10 meters away. We need to find the initial speed, and then find two angles that result in a 6-meter range for the same speed.

Step 2: Horizontal and vertical components of velocity

For a projectile, the horizontal and vertical components of the velocity are crucial. At a 45° angle, these components are equal. If \( v_0 \) is the initial speed, both components are \( \frac{v_0}{\sqrt{2}} \).

Step 3: Formula for Range of a Projectile

The range \( R \) of a projectile fired at an angle \( \theta \) with initial velocity \( v_0 \) is given by \( R = \frac{v_0^2 \sin 2\theta}{g} \) where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

Step 4: Solving for initial speed when range is 10m

Set \( \theta = 45^{\circ} \), so \( \sin(2\theta) = \sin(90^{\circ}) = 1 \). Using the range formula, solve \( 10 = \frac{v_0^2}{9.81} \), leading to \( v_0^2 = 10 \times 9.81 \). Thus, \( v_0 = \sqrt{98.1} \approx 9.90 \) m/s.

Step 5: Solving for angles when range is 6m

Keeping \( v_0 \approx 9.90 \) m/s, use the range formula \( 6 = \frac{(9.90)^2 \sin(2\theta)}{9.81} \). Solving gives \( \sin(2\theta) = \frac{6 \times 9.81}{98.1} = 0.6 \). Calculate \( 2\theta \) using \( \theta = \frac{1}{2} \sin^{-1}(0.6) \).

Step 6: Finding two firing angles

Solve \( 2\theta = \sin^{-1}(0.6) \). The two angles are: \( 2\theta = 36.87^{\circ} \) and \( 2\theta = 180^{\circ} - 36.87^{\circ} = 143.13^{\circ} \). Thus, \( \theta_1 = 18.44^{\circ} \) and \( \theta_2 = 71.56^{\circ} \).

Key Concepts

Initial Speed

In projectile motion, the initial speed, commonly denoted as \( v_0 \), is the speed at which an object is launched. It is crucial because it helps determine how far and how high the projectile will go. When a golf ball is fired from a spring gun, as mentioned in the example, the initial speed must be calculated to predict its trajectory accurately. In this case, the initial speed was calculated by using the range formula.The range formula for a projectile is given by:\[ R = \frac{v_0^2 \sin 2\theta}{g} \]Here:- \( R \) is the range of the projectile,- \( \theta \) is the firing angle,- \( v_0 \) is the initial speed,- \( g \) is the acceleration due to gravity, typically 9.81 m/s².In the scenario where the angle is 45 degrees and the range is 10 meters, the initial speed calculation simplifies a little since \( \sin(90^{\circ}) = 1 \), leading to the equation:\[ 10 = \frac{v_0^2}{9.81} \]Solving this gives \( v_0 = \sqrt{98.1} \approx 9.90 \) m/s. Knowing the initial speed helps us predict the projectile's path.

Firing Angle

The firing angle of a projectile greatly affects its range and height. In projectile motion, this is the angle at which the object, like the golf ball mentioned in the exercise, is launched relative to the horizontal. The angle is essential for determining the components of the initial velocity, which eventually influence how far the projectile will travel.In our example, the golf ball is originally launched at an angle of 45 degrees, which is the optimal angle for achieving maximum horizontal distance for a given speed and assuming a flat launch and landing surface. When dealing with projectile motion, the angle impacts two primary components:- **Horizontal Component**: Dictates how far the projectile travels. It is calculated as \( \frac{v_0}{\sqrt{2}} \) when the angle is 45 degrees.- **Vertical Component**: Determines the height reached. It is also \( \frac{v_0}{\sqrt{2}} \) at 45 degrees.For a different condition where the range needed is only 6 meters, the solutions involve calculating two different angles that satisfy the new range requirement. These angles are found using the range formula and the property that there are generally two complementary angles providing the same range, except for a 45-degree angle. In our example, they were calculated as approximately 18.44 degrees and 71.56 degrees.

Range of a Projectile

The range of a projectile refers to the horizontal distance it travels while in motion, before returning to the same vertical level from which it was launched. It depends on several factors:- **Initial Speed (\( v_0 \))**: Higher initial speeds generally increase the range.- **Firing Angle (\( \theta \))**: The angle determines the distribution between the vertical and horizontal components which impacts how far the projectile will travel.- **Gravity (\( g \))**: It is usually constant (approximately 9.81 m/s²), but it affects how quickly the projectile falls back to the ground.By using the formula \( R = \frac{v_0^2 \sin 2\theta}{g} \), one can calculate the range for a projectile given the initial speed and firing angle. The computation becomes important especially when needing to achieve a specific range, like when adjusting from a 10-meter range to a 6-meter range in the exercise.For example, by maintaining the initial speed and adjusting the angle, such as in our exercise, the golfer can hit targets at different distances. Calculating the range ensures accurate predictions of where the projectile will land. Understanding and utilizing these concepts can aid in strategic planning for games or practical applications in various fields.