Problem 203 - Elementary Algebra

Question

In the following exercises, solve. Round answers to the nearest tenth. An arrow is shot vertically upward from a platform 45 feet high at a rate of \(168 \mathrm{ft} / \mathrm{sec}\). Use the \(\quad\) quadratic \(\quad\) equation \(h=-16 t^{2}+168 t+45\) to find how long it will take the arrow to reach its maximum height, and then find the maximum height.

Step-by-Step Solution

Answer

The arrow takes 5.2 seconds to reach its maximum height of 486.0 feet.

Step 1 - Identify the quadratic equation

The height of the arrow as a function of time t is given by the equation: \[ h = -16t^2 + 168t + 45 \]

Step 2 - Find the time to reach maximum height

The maximum height of a projectile is reached at the vertex of the parabola. For a quadratic equation of the form \[ h = at^2 + bt + c \], the time to reach the vertex (maximum height) is found using the formula: \[ t = -\frac{b}{2a} \]. Here, \( a = -16 \) and \( b = 168 \). So, \[ t = -\frac{168}{2 \cdot (-16)} = \frac{168}{32} = 5.25 \text{ seconds} \].

Step 3 - Calculate the maximum height

Substitute \( t = 5.25 \) back into the original equation to find the maximum height: \[ h = -16(5.25)^2 + 168(5.25) + 45 \]. First, calculate \( (5.25)^2 = 27.5625 \). Then, \(-16 \cdot 27.5625 = -440.9975 \). Next, \( 168 \cdot 5.25 = 882 \). Finally, \[ h = -440.9975 + 882 + 45 = 486.0025 \]. Rounded to the nearest tenth, the maximum height is \( 486.0 \text{ feet} \).

Key Concepts

Maximum Height

The concept of maximum height is crucial when analyzing projectile motion, especially when using quadratic equations. In this exercise, the equation given is \[ h = -16t^2 + 168t + 45 \]. This quadratic equation models the height of an arrow shot vertically upward as a function of time.

The maximum height is the highest point the arrow will reach during its flight. To find this, we need to locate the vertex of the parabolic path represented by the equation.

The vertex is the point where the height is at its peak before it starts falling back down. By substituting the time value obtained from another critical calculation (time to reach the vertex) back into the height equation, we can determine this maximum height accurately.

In this example, substituting \( t = 5.25 \) seconds into the equation gives us a maximum height of 486.0 feet. This simple yet powerful method allows us to understand the peak position of the arrow in its path.

Time to Reach Vertex

Understanding the time to reach the vertex is essential for solving quadratic equations involving projectile motion. This time represents the moment when the projectile (in this case, an arrow) reaches its highest point before descending.

For the quadratic equation \[ h = -16t^2 + 168t + 45 \], the formula to find the time to reach the vertex is \[ t = - \frac{b}{2a} \].

Here, the coefficients are:

• \( a = -16 \)
• \( b = 168 \)

Therefore, substituting these values into the formula, we get: \[ t = - \frac{168}{2 \cdot (-16)} = \frac{168}{32} = 5.25 \]

So, it takes 5.25 seconds for the arrow to reach its vertex or maximum height. This calculation is integral to understanding how long it takes for the arrow to reach the peak of its parabolic flight.

Projectile Motion

Projectile motion is a form of motion experienced by an object thrown or projected into the air, subject to only the acceleration of gravity. The arrow in this exercise exemplifies projectile motion by moving along a curved path due to initial upward velocity and the downward force of gravity.

The quadratic equation \[ h = -16t^2 + 168t + 45 \] is used to describe the height of the projectile at any given time \( t \). The general form of such equations is \[ h = at^2 + bt + c \], where:

• \( a \) is the coefficient of the quadratic term, representing the effect of gravity (usually negative, indicating downward acceleration).
• \( b \) is the coefficient of the linear term, representing the initial upward velocity.
• \( c \) is the constant term, representing the initial height from which the projectile is launched.

The negative sign in front of the \( 16t^2 \) term in our equation indicates gravity's pull affecting the upward motion of the arrow.

Solving for the projectile's maximum height and time to reach the vertex with this equation helps us understand and predict the projectile's motion and behavior throughout its flight. Learning these concepts ensures a deeper understanding of how forces and motion interact in the real-world scenarios.