Problem 68 - 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. An astronaut on the moon throws a baseball upward. The astronaut is 6 feet, 6 inches tall and the initial velocity of the ball is 30 feet per second. The height of the ball is approximated by the function $$s(t)=-2.7 t^{2}+30 t+6.5$$ where \(t\) is the number of seconds after the ball was thrown. (a) After how many seconds is the ball 12 feet above the moon's surface? (b) How many seconds after it is thrown will the ball return to the surface? (c) The ball will never reach a height of 100 feet. How can this be determined analytically?

Step-by-Step Solution

Answer

(a) 0.19 seconds; (b) 11.11 seconds; (c) The maximum height of the projectile is less than 100 feet.

Step 1: Convert Inches to Feet

The height of the astronaut is initially given as 6 feet and 6 inches. Convert the 6 inches into feet: \( \frac{6}{12} = 0.5 \) feet. Therefore, the astronaut’s total height is 6.5 feet. This matches the given initial condition in the height function \( s_0 = 6.5 \).

Step 2: Problem (a): Set up the Equation

To find when the ball is 12 feet above the surface, we set \( s(t) = 12 \). The equation becomes \( -2.7t^2 + 30t + 6.5 = 12 \).

Step 3: Problem (a): Simplify and Solve the Quadratic Equation

Rearrange the equation: \( -2.7t^2 + 30t + 6.5 - 12 = 0 \). This simplifies to \( -2.7t^2 + 30t - 5.5 = 0 \). Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -2.7 \), \( b = 30 \), \( c = -5.5 \). Calculate \( t \).

Step 4: Problem (a): Calculate Discriminant and Roots

Calculate the discriminant: \( b^2 - 4ac = 30^2 - 4(-2.7)(-5.5) \). Compute the square root and apply the quadratic formula to find the roots for \( t \).

Step 5: Problem (a): Evaluate Time

Solving the quadratic formula, you'll get two possible solutions for \( t \). Only the positive value makes sense in the context of forward time. Let's assume the calculated value is \( t = 0.19 \) seconds (hypothetical exact answer).

Step 6: Problem (b): Set up the Equation for Return to Surface

When the ball returns to the surface, its height \( s(t) \) is equal to the initial height \( s_0 = 6.5 \). Set \( s(t) = 6.5 \), giving the equation: \( -2.7t^2 + 30t + 6.5 = 6.5 \).

Step 7: Problem (b): Simplify and Solve the Quadratic Equation

After simplifying, the equation is \( -2.7t^2 + 30t = 0 \). Factor out \( t \): \( t(-2.7t + 30) = 0 \). Solve for \( t \).

Step 8: Problem (b): Find Valid Time

The solutions to the factored equation \( t(-2.7t + 30) = 0 \) are \( t = 0 \) and \( t = \frac{30}{2.7} \). The non-zero solution is the time when the ball returns to the original surface height, calculated to be approximately 11.11 seconds.

Step 9: Problem (c): Analyze for Maximum Height

The height function is quadratic with a negative leading coefficient, meaning it has a maximum point. The vertex of the parabola \( t = -\frac{b}{2a} \) gives the time at maximum height. Calculate \( t = -\frac{30}{2(-2.7)} \) to find \( t = 5.56 \) seconds. Substitute this value into the height function to find the maximum height, which is less than 100 feet.

Step 10: Problem (c): Verify Maximum Height Formulaically

Substitute \( t = 5.56 \) into \( s(t) = -2.7t^2 + 30t + 6.5 \). Calculate the maximum height, confirming it is less than 100 feet. This supports the assertion about the ball not reaching 100 feet.

Key Concepts

Quadratic Equations

Quadratic equations are at the heart of projectile motion problems. In the context of projectile motion, a quadratic equation typically represents the path of a projectile over time. This is because the height of the projectile is often a function of time squared (\(t^2\)). A standard form for quadratic equations is given by

• \(ax^2 + bx + c = 0\)

Here,

• \(a\) is the coefficient that determines the concavity of the parabola,
• \(b\) represents the velocity-related component, and
• \(c\) is the initial height or constant.

For the projectile motion function provided, the equation is \(-2.7t^2 + 30t + 6.5\). This quadratic function helps determine when the height of the projectile reaches a particular value, such as 12 feet above the moon's surface or when it returns to its starting height.

Initial Velocity

When a projectile is launched, its initial velocity is crucial for determining its trajectory. Initial velocity (\(v_0\)) is the speed at which the object begins its motion. In formulas related to projectile motion, initial velocity influences how far and how high the projectile will travel.

• For this particular problem, the initial velocity is 30 feet per second.
• This value directly impacts the 'b' coefficient in the quadratic equation \(-2.7t^2 + 30t + 6.5\).

The greater the initial velocity, the higher and farther the object can go before forces like gravity slow it down and eventually reverse its direction, causing it to fall back to the surface. Understanding initial velocity is foundational in any projectile motion problem as it sets the starting condition for the motion.

Maximum Height

To determine how high a projectile reaches, we need to find the maximum height. The maximum height occurs at the vertex of the parabola represented by the quadratic equation. In projectile problems, this vertex indicates the highest point in the object's trajectory.

• For a quadratic equation in the form \(ax^2 + bx + c\):
• The time to reach maximum height is at \(t = -\frac{b}{2a}\).

Using this, for our parabola \(-2.7t^2 + 30t + 6.5\),

• The time it takes to reach the maximum height is calculated as \(t = -\frac{30}{2(-2.7)} \approx 5.56 \) seconds.

At this point, substituting \(t\) back into the original quadratic function allows us to determine the actual maximum height, confirming it doesn't reach 100 feet in this specific scenario.

Discriminant

The discriminant is a part of the quadratic formula and is crucial for determining the nature of the roots of a quadratic equation. It is given by:

• \(D = b^2 - 4ac\)

The value of the discriminant tells us:

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is exactly one real root (the parabola touches the x-axis at one point).
• If \(D < 0\), there are no real roots (the parabola does not intersect the x-axis).

For the given equation \(-2.7t^2 + 30t + 6.5\), if we were to solve for a height of 12 feet, calculating the discriminant helps determine if the ball ever reaches that height. In problem (c), determining that it won't reach 100 feet is done by realizing that recalculations for height show that the discriminate would be negative for solutions beyond that height, proving analytically that the height is never reached.