Problem 65 - A Graphical Approach to College Algebra

Question

The formula for the height of a projectile is $$s(t)=-16 t^{2}+v_{0} t+s_{0}$$ where \(t\) is time in seconds, \(s_{0}\) is the initial height in feet, \(v_{0}\) is the initial velocity in feet per second, and \(s(t)\) is in feet. Use this formula to solve. A rock is launched upward from ground level with an initial velocity of 90 feet per second. Let \(t\) represent the amount of time elapsed after it is launched. (a) Explain why \(t\) cannot be a negative number in this situation. (b) Explain why \(s_{0}=0\) in this problem. (c) Give the function \(s\) that describes the height of the rock as a function of \(t\) (d) How high will the rock be 1.5 seconds after it is launched? (e) What is the maximum height attained by the rock? After how many seconds will this happen? Determine the answer analytically and graphically. (f) After how many seconds will the rock hit the ground? Determine the answer graphically.

Step-by-Step Solution

Answer

(a) Time, \( t \), cannot be negative. (b) The rock starts at ground level, so \( s_0 = 0 \). (c) \( s(t) = -16t^2 + 90t \). (d) 99 feet after 1.5 seconds. (e) Max height of 126.56 feet at \( t = 2.8125 \) seconds. (f) Hits ground at \( t = 5.625 \) seconds.

Step 1: Understanding Time Constraint (Part a)

In this problem, time \( t \) represents the duration after the rock is launched. Time cannot be negative because it measures elapsed time starting from the launch moment. Negative time would imply moments before the launch event, which is not possible for this situation.

Step 2: Initial Height Explanation (Part b)

The rock is launched from ground level, meaning its starting position is at height zero. Therefore, the initial height \( s_0 \) is 0.

Step 3: Write the Height Function (Part c)

Using the formula \( s(t) = -16t^2 + v_0t + s_0 \), and substituting \( v_0 = 90 \) feet per second and \( s_0 = 0 \), we have:\[s(t) = -16t^2 + 90t\]This is the function that describes the height of the rock over time.

Step 4: Calculate Height at 1.5 Seconds (Part d)

Substitute \( t = 1.5 \) seconds into the height function:\[s(1.5) = -16(1.5)^2 + 90(1.5)\]First, calculate \( (1.5)^2 = 2.25 \), then:\[s(1.5) = -16 imes 2.25 + 90 imes 1.5 = -36 + 135 = 99 ext{ feet}\]The rock is 99 feet high after 1.5 seconds.

Step 5: Determine Maximum Height and Time (Part e)

The maximum height occurs at the vertex of the parabola described by the function. The time \( t \) to reach this maximum height can be found using \( t = -\frac{b}{2a} \) from the vertex formula for quadratics:\[t = - \frac{90}{2(-16)} = \frac{90}{32} = 2.8125 ext{ seconds}\]Substitute back into the height function:\[s(2.8125) = -16(2.8125)^2 + 90(2.8125)\]Calculate \( (2.8125)^2 = 7.91015625 \), then:\[s(2.8125) = -16 imes 7.91015625 + 90 imes 2.8125 = -126.5625 + 253.125 = 126.5625 ext{ feet}\]The maximum height achieved by the rock is approximately 126.56 feet, which occurs at 2.8125 seconds.

Step 6: Find Time to Hit Ground (Part f)

The rock hits the ground when \( s(t) = 0 \). Solve \( -16t^2 + 90t = 0 \):\[-16t(t - \frac{90}{16}) = 0 \]The solutions are \( t = 0 \) or \( t = \frac{90}{16} = 5.625 \) seconds.Since \( t = 0 \) is the launch time, the correct answer is around \( 5.625 \) seconds. This conclusion can also be verified by graphically plotting the function and identifying when \( s(t) = 0 \) after the launch.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function where the highest degree of the variable is squared. It can be written in the standard form: \( ax^2 + bx + c \). In projectile motion problems, the quadratic function is used to describe the path or trajectory of an object being thrown into the air. This function determines the height of the object over time. The coefficients \( a \), \( b \), and \( c \) will change depending on factors like initial speed and height. In physics, particularly in projectile motion, the function takes the form \( s(t) = -16t^2 + v_0t + s_0 \), where \( t \) is time, \( v_0 \) is initial velocity, and \( s_0 \) is initial height. The

• \( -16 \) represents the acceleration due to gravity in feet per second squared,
• \( v_0 \) is the initial upward velocity, and
• \( s_0 \) denotes the starting height of the projectile.

This setup creates a parabolic path typical of projectile motion, allowing us to predict the object's position at any given time.

Vertex of a Parabola

The vertex of a parabola is a critical point representing the highest or lowest point on the graph of a quadratic function, depending on the parabola's direction. In the context of projectile motion, the vertex corresponds to the maximum height that the object reaches.To find the vertex analytically, we use the vertex formula \( t = -\frac{b}{2a} \), where \( a \) and \( b \) are the coefficients from the quadratic function \( ax^2 + bx + c \). In our context, the maximum height is achieved when the vertical velocity is zero due to the gravitational pull.By locating the vertex, you understand when and how high the projectile will peak before descending. This point gives us vital information needed for applications in sports, physics problems, and even engineering scenarios.

Maximum Height Calculation

The calculation of the maximum height the projectile reaches is directly linked to the vertex of the parabola. To compute this, you first determine the time it takes for the rock to reach its peak height using the vertex formula \( t = -\frac{b}{2a} \). Once you have this time, substitute it back into the original height equation to find the maximum height.In the given problem:

• \( t = \frac{90}{32} = 2.8125 \) seconds,
• the maximum height calculation follows by substituting \( t \) back into the function: \( s(2.8125) = -16(2.8125)^2 + 90(2.8125) \).
• Here, the projectile's highest point, or maximum height, is approximately 126.56 feet.

This phase of calculation allows you to evaluate the maximum potential energy of the projectile during its flight, a vital piece of analysis, especially in ballistic testing and other practical applications.

Time of Flight in Projectile Motion

The time of flight in projectile motion refers to the total time the projectile spends in the air from launch until it returns to the ground. In mathematical terms, this is the duration until the height function returns to zero after it has been released.For quadratic functions related to projectile motion, solving for the time of flight means setting the height equation to zero \( s(t) = 0 \) and finding the non-zero solution. This identifies when the rock will return to ground level.From the problem, the function \( -16t^2 + 90t = 0 \) is solved by factoring or using the quadratic formula. This gives us:

• Time to hit the ground at \( t = \frac{90}{16} = 5.625 \) seconds.

Understanding the time of flight is crucial for determining the range and coverage of projectile during its journey, impacting various fields such as military defense systems, sports dynamics, and space exploration.