Problem 48 - Calculus Single Variable

Question

Concern an object that is propelled straight up. Its height at time \(t\) seconds is given in feet by \(H(t)=-16 t^{2}+128 t+68\). For how many seconds does the object rise?

Step-by-Step Solution

Answer

The object rises for 4 seconds.

Step 1: Understand the problem

We are given the height function \(H(t) = -16t^2 + 128t + 68\), where \(H(t)\) represents the height in feet at time \(t\) seconds. The goal is to find how long the object rises.

Step 2: Determine the vertex of the parabola

The function \(H(t) = -16t^2 + 128t + 68\) is a quadratic equation representing a downward-opening parabola. The object rises until it reaches its maximum height at the vertex. The vertex of the parabola \(ax^2 + bx + c\) is at \(t = \frac{-b}{2a}\).

Step 3: Calculate the time at maximum height

Use the formula for the vertex: \(t = \frac{-b}{2a}\), where \(a = -16\) and \(b = 128\). Substitute the values:\[t = \frac{-128}{2(-16)} = \frac{-128}{-32} = 4.\]The object reaches its maximum height at \(t = 4\) seconds.

Step 4: Formulate the answer

Since the object rises until it reaches its maximum height, it rises for 4 seconds.

Key Concepts

Parabola

A parabola is a U-shaped graph that represents a quadratic function. When you graph a quadratic equation like \( H(t) = -16t^2 + 128t + 68 \), the shape of the graph is a parabola.
For this specific equation, the parabola opens downward because the coefficient of the \( t^2 \) term is negative (-16). This means as you plug in values for \( t \), the height \( H(t) \) increases up to a certain point and then starts to decrease again.

• In a downward-opening parabola, the highest point you can reach is the vertex.
• Understanding the parabola helps us to know that the projectile will rise until it hits this vertex.

Downward-opening parabolas are common when dealing with problems about objects being thrown or propelled upwards.

Maximum Height

The maximum height of an object propelled upwards is the highest point it reaches, which corresponds to the vertex of its parabolic trajectory. In real-world terms, this is the peak height before the object starts descending back down.
To find the maximum height, it's crucial to identify when this happens in terms of time.

• Once we find the time \( t \) at which the vertex occurs, we can substitute it back into the height function to find this maximum height value.
• In most cases, like our example, it's important to remember that the object's rise ends at this point.

This concept is especially useful in various fields like physics and engineering.

Vertex of a Parabola

The vertex of a parabola is the most important point working with a quadratic function as it signifies either the maximum or the minimum point of the graph.
For upward-opening parabolas, the vertex is the minimum, and for downward-opening parabolas like ours, it is the maximum.
The formula for the vertex of a parabola, given by the quadratic \( ax^2 + bx + c \), is \( t = \frac{-b}{2a} \).

• In our example \( H(t) = -16t^2 + 128t + 68 \), \( a = -16 \) and \( b = 128 \).
• Using the formula, \( t = \frac{-128}{2(-16)} = 4 \), provides the time for the maximum height.

Finding the vertex helps us determine exactly when the maximum or minimum value occurs.

Height Function

A height function, like \( H(t) = -16t^2 + 128t + 68 \), describes how high an object is at any time \( t \). It’s commonly used in physics to predict the motion of an object shot straight up.
Each component in the height function has a specific meaning:

• The \(-16t^2\) term is the acceleration due to gravity in feet per second squared.
• The \(128t\) term represents the initial velocity; in this case, it's the force propelling the object upwards.
• The constant \(68\) is the starting height or initial position of the object.

By using the height function, one can find out where an object will be at any point in time during its flight.