Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. A golf ball is hit so that its height \(h\) in feet after \(t\) seconds is given by $$h(t)=-16 t^{2}+60 t$$ (a) What is the initial height of the golf ball? (b) How high is the golf ball after 1.5 seconds? (c) Find the maximum height of the golf ball.

Step-by-Step Solution

Answer

(a) 0 feet; (b) 54 feet; (c) 56.25 feet.

Step 1: Determine the Initial Height

The initial height of the golf ball is the height at time \(t = 0\). Substitute \(t = 0\) into the function \(h(t) = -16t^2 + 60t\):\[h(0) = -16(0)^2 + 60(0) = 0.\]Thus, the initial height of the golf ball is 0 feet.

Step 2: Calculate the Height at 1.5 Seconds

Substitute \(t = 1.5\) into the function \(h(t) = -16t^2 + 60t\) to find the height after 1.5 seconds:\[h(1.5) = -16(1.5)^2 + 60(1.5)\]\[= -16(2.25) + 90\]\[= -36 + 90\]\[= 54.\]Therefore, the height of the golf ball after 1.5 seconds is 54 feet.

Step 3: Find the Maximum Height

To find the maximum height, we need to determine the vertex of the parabola represented by the quadratic function, as the vertex will give us the maximum value for a parabola opening downwards. The time at which the maximum height occurs is \[t = \frac{-b}{2a}\]for a quadratic equation of the form \(h(t) = at^2 + bt + c\), where \(a = -16\), \(b = 60\).Substitute into the formula:\[t = \frac{-60}{2(-16)} = \frac{-60}{-32} = \frac{15}{8} = 1.875\]Now, substitute \(t = 1.875\) back into the function to find the maximum height:\[h(1.875) = -16(1.875)^2 + 60(1.875)\]\[= -16(3.515625) + 112.5\]\[= -56.25 + 112.5\]\[= 56.25.\]Thus, the maximum height of the golf ball is 56.25 feet.

Key Concepts

Projectile Motion

Projectile motion is a type of motion experienced by objects that are launched into the air and are subject to the acceleration due to gravity. This type of motion is characterized by the object's path, which is typically a curve called a parabola.

• When an object follows projectile motion, its horizontal and vertical movements are independent.
• The acceleration due to gravity acts to pull the object downward, affecting only its vertical motion.
• This is why a projectile generally follows a curved path instead of a straight line.

For the golf ball in this exercise, its height as a function of time is affected by gravity, which is represented by the quadratic term \(-16t^2\). The initial velocity, on the other hand, represented by the term \(60t\), determines how fast and in which direction the ball was initially hit. Together, these factors define the path and ultimate motion of the projectile.

Parabolas

A parabola is a symmetric, U-shaped curve that can open upwards or downwards. In mathematics, parabolas are the graphs of quadratic functions, which have the general form \(y = ax^2 + bx + c\).

• If \(a > 0\), the parabola opens upward, and if \(a < 0\), it opens downward.
• The value and sign of \(a\) significantly affect the shape and direction of the parabola.
• Each parabola is symmetric around its axis, which makes it easy to identify and describe.

In this case, the quadratic function of the golf ball, \(-16t^2 + 60t\), forms a parabola that opens downward due to the negative coefficient of the \(t^2\) term. This is typical of projectile motion scenarios where gravity pulls objects back down to the ground.

Vertex of a Parabola

The vertex of a parabola is a critical point that represents either the maximum or minimum value of the quadratic function, depending on the direction in which the parabola opens. For a parabola that opens downward, like the golf ball's trajectory, the vertex will be the maximum point.

• The vertex can be found using the formula \(-b/(2a)\), where \(a\) and \(b\) are coefficients of the quadratic function.
• This gives us the \(t\)-value at which the maximum height occurs.
• By substituting this \(t\)-value back into the function, we can determine the maximum height.

In the example given, the vertex occurred at \(t = 1.875\) seconds, resulting in a height of \(56.25\) feet. The process of finding the vertex presents a powerful tool for solving many real-world problems involving parabolic motion.