Problem 105 - Algebra and Trigonometry Real Mathematics, Real People

Question

Consider the path of a projectile projected horizontally with a velocity of \(v\) feet per second at a height of \(s\) feet, where the model for the path is \(x^{2}=-\frac{v^{2}}{16}(y-s)\). In this model (in which air resistance is disregarded), \(y\) is the height (in feet) of the projectile and \(x\) is the horizontal distance (in feet) the projectile travels. A ball is thrown from the top of a 100 -foot tower with a velocity of 28 feet per second. (a) Find the equation of the parabolic path. (b) How far does the ball travel horizontally before striking the ground?

Step-by-Step Solution

Answer

The equation of the parabolic path is \(x^{2} = -49y + 4900\), and the ball travels a distance of 70 feet before hitting the ground.

Step 1: Substituting the given values into the equation

The first step is to take our general equation \(x^{2}=-\frac{v^{2}}{16}(y-s)\) and replace \(v\) and \(s\) by their values:\(x^{2} = -\frac{(28)^{2}}{16}(y - 100) = -49(y - 100) = -49y + 4900\). So, the equation of the parabolic path is \(x^{2} = -49y + 4900\).

Step 2: Computing the horizontal distance

To compute how far the ball travels horizontally before it hits the ground, we need to set the value of \(y\) in our equation to 0 and solve for \(x\):\((-49*0 + 4900) = x^2 \Rightarrow x = \sqrt{4900} = 70\) feet. So, the ball will travel 70 feet horizontally before it hits the ground.

Key Concepts

Parabolic Path

In projectile motion, the path followed by an object is often shaped like a parabola. This is because the object moves both horizontally and vertically under the influence of gravity. The equation of the parabola is derived by eliminating air resistance to make the calculations simpler.

In our exercise, the parabolic path is represented by the equation \(x^{2} = -49y + 4900\). This equation helps us understand how the projectile moves and what shape the path takes. The "\(x\)" represents the horizontal movement, while the "\(y\)" represents the vertical fall.

• Each point \((x, y)\) on this curve tells us the position of the projectile at a particular instant of time.
• The negative sign in the path equation indicates that as the projectile moves forward, it also falls downwards due to gravity.

This simple yet effective mathematical representation allows us to predict the course of the projectile effectively.

Horizontal Distance

One of the primary concerns with projectile motion is how far the object will travel horizontally. The horizontal distance, often termed "range", shows the reach or extent of how far the projectile can go before hitting the ground. In this situation, the horizontal distance is calculated using the parabolic equation and setting the vertical component, \(y\), to zero.

For the given problem, the calculation follows as:

• Start with the equation \(x^{2} = -49y + 4900\).
• Set \(y = 0\) to determine when the projectile hits the ground.
• Then, solve for \(x\) to find \(x^{2} = 4900\).
• Finally, we solve \(x = \sqrt{4900} = 70\) feet.

Therefore, the ball travels 70 feet horizontally before hitting the ground. It is important to recognize that this distance depends mainly on the initial velocity and height.

Velocity

Velocity in projectile motion is broken into two components: horizontal and vertical. These components are independent of each other. Horizontal velocity determines how far the projectile travels, while vertical velocity affects how long it stays in the air. In the context of a projectile fired horizontally, the initial velocity \(v\) is crucial to determining the path and the final position.

For our exercise, we used an initial horizontal velocity of 28 feet per second.

• The impact of this velocity is seen directly in the formula \(x^{2} = -\frac{v^{2}}{16}(y - s)\),
• where it ultimately calculated the projectile's reach by altering the structure of the parabola.

This velocity remains constant throughout the motion because we disregard air resistance. Consistent velocity ensures a symmetrical and predictable parabolic path.

Height

Height plays a vital role in the motion of a projectile as it determines the starting point and helps calculate the time it takes for the projectile to hit the ground.

In this exercise, the projectile starts from a tower at a height \(s\) of 100 feet. The equation \(x^{2} = -49y + 4900\) incorporates initial height by the term \((y - s)\).

• The greater the initial height, the longer the projectile remains in the air.
• This extended time allows the object to travel a greater horizontal distance before landing.
• Height directly influences the downward component of the parabolic path.

By initiating the movement from a higher point, the projectile gets more travel time due to gravity's pull, resulting in wider coverage of horizontal distance.