Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object thrown directly upward from ground level with an initial velocity of 48 feet per second is \(s=48 t-16 t^{2}\) feet high at the end of \(t\) seconds. (a) What is the maximum height attained? (b) How fast is the object moving, and in which direction, at the end of 1 second? (c) How long does it take to return to its original position?

Step-by-Step Solution

Answer

(a) 36 feet; (b) 16 ft/s upward; (c) 3 seconds.

Step 1: Find the Time to Reach Maximum Height

To determine the maximum height, we first find when the object reaches it. The maximum height occurs when the velocity is zero. Calculate the derivative of the height function, which gives the velocity: \( v(t) = \frac{d}{dt}(48t - 16t^2) = 48 - 32t \).Set the velocity to zero and solve for \( t \):\[0 = 48 - 32t \32t = 48 \t = \frac{48}{32} = 1.5 \, \text{seconds}.\]

Step 2: Calculate Maximum Height

Substitute the time when the object reaches its maximum height back into the height equation:\[ s = 48(1.5) - 16(1.5)^2 \].Simplify the expression:\[48 \cdot 1.5 = 72 \16 \cdot 1.5^2 = 36 \s = 72 - 36 = 36 \, \text{feet}.\]

Step 3: Determine the Velocity at 1 Second

Evaluate the velocity function at \( t = 1 \) second:\[v(t) = 48 - 32t \v(1) = 48 - 32 \cdot 1 = 16 \, \text{feet per second}.\]The object is moving upward since the velocity is positive.

Step 4: Calculate Time to Return to Original Position

To find when the object returns to its original position (ground level), set the height function to zero and solve for \( t \):\[ s = 48t - 16t^2 = 0 \0 = t(48 - 16t) \t = 0 \quad \text{or} \quad 48 - 16t = 0 \16t = 48 \t = \frac{48}{16} = 3 \, \text{seconds}.\]The object returns to the ground at \( t = 3 \) seconds.

Key Concepts

Maximum Height

To find the maximum height of an object projected upward, you need to know when it stops moving higher. This moment is signified by its velocity reaching zero.
For the problem shown, the height function is given as \( s = 48t - 16t^2 \). To determine the time when the maximum height occurs, calculate the derivative of this function, since the derivative represents velocity.

• The velocity function derived is \( v(t) = 48 - 32t \).
• Set the velocity to zero: \( 48 - 32t = 0 \).
• Solve for \( t \) to get \( t = 1.5 \) seconds.

At \( t = 1.5 \) seconds, substitute back into the height equation to find the maximum height:

• \( s = 48(1.5) - 16(1.5)^2 \).
• This simplifies to \( 72 - 36 \).
• Maximum height is \( 36 \) feet.

Initial Velocity

When an object is thrown upward, it starts with an initial velocity, which can be thought of as the speed with which it is launched.
In this equation, you can identify the initial velocity from the height function. Here, the initial velocity is the coefficient of the \( t \) term in the equation \( s = 48t - 16t^2 \).

• The initial velocity is \( 48 \) feet per second.
• This is the speed of the object at \( t = 0 \).

Understanding the initial velocity helps in analyzing the object’s motion, such as calculating how quickly it will rise and when it will start falling back.

Derivative

In calculus, a derivative represents the rate of change of a function. In motion problems, derivatives are particularly useful, helping us determine velocity or acceleration from a position function.
By deriving the height equation \( s = 48t - 16t^2 \), we get the velocity function.

The Derivative Process:

• Start with \( s = 48t - 16t^2 \).
• Find \( v(t) = \frac{d}{dt}(48t - 16t^2) = 48 - 32t \).

This process shows how quickly the object's height changes over time and helps pinpoint when it reaches the maximum height. Derivatives make predictions about future motions possible, too.

Quadratic Equations

Quadratic equations are polynomials of degree two, often taking the form \( ax^2 + bx + c = 0 \). In motion problems like this, they model the path of an object under uniform acceleration.
The height equation \( s = 48t - 16t^2 \) is a quadratic equation.

Why Quadratics Matter:

• They describe the parabolic path of an object.
• Solving the equation reveals key moments, like when the object reaches the ground again.

In this exercise, to find when the object returns to ground level, set \( s = 0 \) and solve \( 48t - 16t^2 = 0 \) using methods like factoring:

• This gives solutions \( t = 0 \) (the start) or \( t = 3 \) seconds (when it returns to ground).

Such equations are powerful tools in predicting the behavior of objects in free-fall or projectile scenarios.