Problem 97 - Algebra and Trigonometry

Question

Falling-Body Problems Use the formula \(h=-16 t^{2}+v_{0} t\) discussed in Example 7. A ball is thrown straight upward at an initial speed of \(v_{0}=40 \mathrm{ft} / \mathrm{s}\) (a) When does the ball reach a height of 24 \(\mathrm{ft}\) ? (b) When does it reach a height of 48 \(\mathrm{ft}\) ? (c) What is the greatest height reached by the ball? (d) When does the ball reach the highest point of its path? (e) When does the ball hit the ground?

Step-by-Step Solution

Answer

The ball reaches 24 ft at 1 and 1.5 s, 48 ft at 1 and 3 s, and the greatest height of 25 ft at 1.25 s. It hits the ground at 2.5 s.

Step 1: Identify Known Values

We know the initial speed \(v_0\) of the ball is 40 ft/s. The height function is \(h = -16t^2 + 40t\). We will use this equation to find heights at different times.

Step 2: Set up the Equation for (a)

To find when the ball reaches a height of 24 ft, we plug \(h = 24\) into the equation: \(-16t^2 + 40t = 24\).

Step 3: Solve the Quadratic for (a)

Rearrange to form the quadratic equation \(-16t^2 + 40t - 24 = 0\). Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \(a = -16, b = 40, c = -24\).

Step 4: Calculate Discriminant for (a)

Calculate the discriminant \(b^2 - 4ac\) which is \(40^2 - 4(-16)(-24) = 1600 - 1536 = 64\).

Step 5: Find Solutions for (a)

Apply the quadratic formula to find \(t = \frac{-40 \pm \sqrt{64}}{-32}\), which simplifies to \(t = 1\) and \(t = 1.5\). The ball reaches 24 ft at 1 second and again at 1.5 seconds.

Step 6: Set up the Equation for (b)

Similarly, for a height of 48 ft, set \(h = 48\) leading to \(-16t^2 + 40t = 48\). Rearrange to form \(-16t^2 + 40t - 48 = 0\).

Step 7: Solve the Quadratic for (b)

Again use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = -16, b = 40, c = -48\).

Step 8: Calculate Discriminant for (b)

\(b^2 - 4ac = 40^2 - 4(-16)(-48) = 1600 - 3072 = 64\).

Step 9: Find Solutions for (b)

Solve for \(t\), \(t = \frac{-40 \pm \sqrt{64}}{-32}\), leading to \(t = 1\) and \(t = 3\). The ball reaches 48 ft at 1 second and again at 3 seconds.

Step 10: Determine Maximum Height (c)

To find the maximum height, recognize that it occurs at the vertex of the parabola \(h = -16t^2 + 40t\). Use \(t = -b/(2a)\) where \(a = -16\) and \(b = 40\): \(t = -40/(2 \times -16) = 1.25\)s.

Step 11: Calculate Maximum Height (c)

Substitute \(t = 1.25\) into the height equation: \(h = -16(1.25)^2 + 40(1.25)\) yields approximately \(25\) ft. The greatest height is 25 feet.

Step 12: Time of Greatest Height (d)

Use the result from the previous step: the ball reaches its greatest height at \(t = 1.25\) seconds.

Step 13: Determine Time to Hit Ground (e)

Set \(h = 0\) to find when the ball hits the ground: \(-16t^2 + 40t = 0\). Factor to get \(t(-16t + 40) = 0\) leading to solutions \(t = 0\) and \(t = 2.5\). Thus, it hits the ground at 2.5 seconds.

Key Concepts

Quadratic Formula

One of the most versatile tools in solving quadratic equations is the Quadratic Formula. This formula helps us find the values of a variable that satisfy a quadratic equation, which is in the form of \(ax^2 + bx + c = 0\). To solve it, use the formula:

• \( t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)

Here, \(a\), \(b\), and \(c\) are constants from the equation, "\(ax^2 + bx + c\)". The \(b^2 - 4ac\) part inside the square root is known as the discriminant and it determines the nature of the roots. Sometimes, students face difficulty initially, but with practice, this formula becomes a straightforward method to extract roots for any quadratic equation. It allows for quick adjustments depending on the values for heights at certain times, as seen in falling body problems.

Maximum Height Calculation

Finding the maximum height reached by an object on a parabolic path is a practical application of calculus concepts in conjunction with kinematics. In the specific case of a ball thrown upwards, the equation often used is

• \(h = -16t^2 + v_0t\)

where \(h\) represents the height, and \(t\) time. To get maximum height, remember it's at the vertex of the parabola. Mathematically, calculate the time at which the height is at its maximum using

• \(t = -\frac{b}{2a}\)

Plug this time back into the original equation to find the highest point the object reaches. It's the combination of mechanics and algebra, providing exciting insights into the world of motion!

Parabolic Motion

When an object is projected with a certain initial velocity, its motion forms a parabolic trajectory. This is described by quadratic equations. For a ball thrown vertically, the motion is a simple parabola opening downward due to gravity. Intuitively, you can expect the ball to go up, slow down, stop at the peak (highest point), and then accelerate downward until reaching the ground. This motion can be represented by

• \(h = -16t^2 + v_0t\)

The quadratic nature of this equation ensures that the object's path will be a symmetrical curve. Understanding this concept allows us to predict the entire motion of the ball, including when and where it will be at any point in time, such as when it will reach a certain height or hit the ground again.

Discriminant Calculation

In quadratic equations, the discriminant provides crucial information about the nature of the roots. The discriminant is found by evaluating

• \(b^2 - 4ac\)

If this value is positive, you'll have two real roots, indicating two different times the object reaches the specific height. If it's zero, there's exactly one real root, meaning the object touches the height only once. A negative discriminant implies no real roots; the height is never reached by the object in its motion. For falling body problems, the discriminant helps determine the times at which the ball reaches specific heights or when it hits the ground. This concept transforms potential confusion into clarity, simplifying complex quadratic calculations and enhancing your problem-solving skills.