Question

Use the model for projectile motion, assuming there is no air resistance. A bale ejector consists of two variable-speed belts at the end of a baler. Its purpose is to toss bales into a trailing wagon. In loading the back of a wagon, a bale must be thrown to a position 8 feet above and 16 feet behind the ejector. (a) Find the minimum initial speed of the bale and the corresponding angle at which it must be ejected from the baler. (b) The ejector has a fixed angle of \(45^{\circ} .\) Find the initial speed required for a bale to reach its target.

Short answer

For part (a), the minimum initial speed is a value derived from the quadratic derived from the equation of trajectory, and the corresponding angle is the one that results in this speed. For part (b), the required speed with an ejection angle of 45 degrees can be calculated by rearranging the range equation.

Step-by-step solution

Identify Given Quantities

From the problem, the horizontal distance (x) is 16ft and the vertical distance (y) is 8ft. The acceleration due to gravity (g) is approximately 32.2ft/s². These are the quantities needed for the problem.

Minimum Initial Speed and Corresponding Ejection Angle

The equation of trajectory for a projectile motion is \[ \tan(\theta) = \frac{y}{x} + \frac{g x}{2 v^2 \cos^2(\theta)} \] Simplifying the equation gives \[ v^2 = \frac{g x (x + \sqrt{x^2 + 4 y h})}{2 x^2 \cos^2(\theta)} \] where h is the height that the projectile must reach. Solve for v which will give a quadratic in terms of \( \cos(\theta) \). Optimizing the solution for minimum speed will give this as the required speed and corresponding initial ejection angle.

Initial Speed for 45 Degrees Angle

As we identified earlier, the range of a projectile is given by \[ R = \frac{v^2 sin(2\theta)}{g} \]. Setting R equal to x (the given horizontal distance) will allow us to solve for the initial speed (v).

Key concepts

Initial Speed in Projectile Motion

Understanding the initial speed is crucial for solving projectile motion problems. It is the speed at which an object is launched, and it greatly determines the trajectory of the object. In the case of the bale ejector, finding the minimum initial speed is necessary to ensure the bale reaches a position 8 feet above and 16 feet behind the ejector.

To solve for the initial speed, we first look at the equation that relates horizontal distance, vertical distance, and the gravitational force. The initial speed will need to be sufficient to overcome gravity and reach the required distance. Algebraically, when solving for the speed in this context, we end up with a quadratic equation due to the square of the initial speed in the trajectory equation. The smallest positive root of this equation will provide us with the minimum initial speed necessary for the bale to reach its target.

For students to better grasp this concept, it's helpful to consider that the higher the initial speed, the farther and higher an object can potentially travel, assuming the angle of ejection and other conditions remain constant.

Trajectory Equation

The trajectory equation is essential in understanding the path of a projectile. This equation incorporates the initial speed, the angle of ejection, the horizontal distance, and the influence of gravity on the object's flight.

Mathematically, the trajectory of a projectile is described by a parabola in the absence of air resistance. It is derived from the kinematic equations of motion and represents the y-coordinate (vertical position) as a function of the x-coordinate (horizontal position). The general form includes terms for the initial velocity, the angle of release, and gravitational acceleration.

To apply the trajectory equation to our bale ejector problem, one must set up the equation with the given distances and solve for the unknowns, which are the initial speed and the angle of ejection. By doing this, it is possible to predict the path of the bale and ensure that it lands at the specified location.

Ejection Angle

The ejection angle, also known as the launch angle, is the angle at which an object is projected into the air. It has a profound impact on the projectile's range and height. For instance, in our textbook problem, the required ejection angle determines how the bale will be thrown into the wagon.

An angle of 45 degrees maximizes the range of the projectile under ideal conditions. However, when a specific target point is given as in the case of the bale and wagon, the optimal ejection angle must be calculated to ensure the bale follows a trajectory that will lead to the desired landing spot.

Using trigonometric functions as part of the trajectory equation, we solve for the ejection angle that meets our conditions. This problem-solving exercise illustrates how different ejection angles influence the trajectory and how they must be adjusted according to the task at hand. Moreover, understanding this concept enables us to control projectile motion with greater precision, which is key in various real-world applications beyond just this example.