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 feet/second upward, (c) 3 seconds.

Step 1: Determine the Time to Reach Maximum Height

The object reaches the maximum height when the velocity, which is the derivative of the height function, is zero. The height function is given by \(s = 48t - 16t^2\). The velocity \(v(t)\) is the derivative of \(s\), so \(v(t) = s'(t) = 48 - 32t\). Set \(v(t) = 0\) to find \(t\): \(48 - 32t = 0\) which gives \(t = \frac{3}{2}\) or 1.5 seconds.

Step 2: Calculate Maximum Height

Substitute \(t = 1.5\) (from Step 1) into the height equation \(s = 48t - 16t^2\) to find \(s(1.5)\):\[s(1.5) = 48(1.5) - 16(1.5)^2\]\[= 72 - 36 = 36\]Thus, the maximum height attained is 36 feet.

Step 3: Find Velocity at t = 1 Second

The velocity function is \(v(t) = 48 - 32t\). Substitute \(t = 1\) into \(v(t)\): \[v(1) = 48 - 32(1) = 16\]This means the object is moving upward at 16 feet per second at the end of 1 second.

Step 4: Calculate Time to Return to Ground

To find when the object returns to its original position (ground level), solve for \(t\) when \(s = 0\):\[48t - 16t^2 = 0\]Factor the equation: \(t(48 - 16t) = 0\). This gives two solutions: \(t = 0\) (initially at ground) and \(48 - 16t = 0\), so \(t = 3\). Thus, the object returns to the ground after 3 seconds.

Key Concepts

Maximum Height Calculation

To determine how high an object travels when thrown upwards, we must calculate the maximum height it achieves. This happens when the vertical velocity of the object is zero because at this point, it stops rising and begins to fall back down.
To find the maximum height, start by setting the velocity function equal to zero. For this exercise, the velocity function, derived from the height function, is given by \[ v(t) = 48 - 32t \].

• Set this equation equal to zero: \[ 48 - 32t = 0 \].
• Solve for \( t \): \[ t = \frac{3}{2} \] (i.e. 1.5 seconds).

This is the time it takes for the object to reach its peak height.
Next, substitute \( t = 1.5 \) back into the height equation \[ s = 48t - 16t^2 \] to find the maximum height:

• \[ s(1.5) = 48(1.5) - 16(1.5)^2 = 36 \].

Thus, the object reaches a maximum height of 36 feet.

Velocity Function

Understanding how fast the object moves and in which direction requires knowledge of its velocity function. The velocity function is the derivative of the height function.
In this case, the height function is given by \[ s(t) = 48t - 16t^2 \], so the velocity function, derived by differentiating with respect to time \( t \), is \[ v(t) = 48 - 32t \].

• To understand the object's motion at any point in time, simply evaluate \( v(t) \).

For instance, to find the velocity at the end of 1 second, substitute \( t = 1 \) into the velocity function:

• \[ v(1) = 48 - 32 \times 1 = 16 \] feet per second.

This indicates the object is moving upward at 16 feet per second, demonstrating the direct relationship between the velocity function and the object's speed and direction.

Time of Flight

Calculating the time of flight is key to understanding how long it takes for a thrown object to return to the ground. This involves finding when the object's height equation equals zero, indicating it has come back to its initial position.
Given the height function \[ s(t) = 48t - 16t^2 \], we set it to zero: \[ 48t - 16t^2 = 0 \].

• Factor this equation: \[ t(48 - 16t) = 0 \].
• Solve for \( t \), resulting in two solutions: \[ t = 0 \] (initial time), and \[ 48 - 16t = 0 \], giving \[ t = 3 \].

Thus, the object returns to its original position after 3 seconds. This solution signifies the complete journey, going up and then descending back.

Quadratic Functions

Quadratic functions often model scenarios involving projectile motion, like the path of an object thrown upwards.
The general form of a quadratic function is \[ ax^2 + bx + c \]. In this example, the height function is \[ s(t) = 48t - 16t^2 \], matching the structure of a quadratic equation \[ -16t^2 + 48t + 0 \].

• The coefficient \(-16\) indicates the parabola opens downwards, which is typical for projectile motion due to the effect of gravity.
• The maximum point of this parabola represents the maximum height the object reaches.

Solving quadratic equations often involves finding when the function equals zero (ground level), using techniques such as factoring or the quadratic formula.
The properties of quadratic functions provide a deeper insight into the relationships between variables like time, height, and velocity.