Problem 87 - Algebra for College Students

Question

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Tossing a ball. A ball is tossed into the air at \(10 \mathrm{ft} / \mathrm{sec}\) from a height of 5 feet. How long does it take to reach the earth?

Step-by-Step Solution

Answer

0.95 sec

Step 1 - Define the equation

The height h(t) of an object tossed into the air with an initial velocity v from an initial height h0 is given by the equation: \[ h(t) = -\frac{1}{2}gt^2 + vt + h_0 \] where g is the acceleration due to gravity (approximately 32 ft/sec²), v is the initial velocity, and h0 is the initial height. For this problem, v = 10 ft/sec and h0 = 5 ft.

Step 2 - Input the values

Substitute the values into the height equation: \[ h(t) = -\frac{1}{2}(32)t^2 + 10t + 5 \] Simplify: \[ h(t) = -16t^2 + 10t + 5 \]

Step 3 - Set up the equation to solve for time

To find when the ball reaches the earth, set the height h(t) to 0 and solve for t: \[ 0 = -16t^2 + 10t + 5 \]

Step 4 - Solve the quadratic equation

Rearrange the quadratic equation to standard form: \[ -16t^2 + 10t + 5 = 0 \] Use the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where a = -16, b = 10, and c = 5.

Step 5 - Calculate the discriminant

Calculate the discriminant: \[ b^2 - 4ac = 10^2 - 4(-16)(5) = 100 + 320 = 420 \]

Step 6 - Find the solutions using the quadratic formula

Plug the values into the quadratic formula: \[ t = \frac{-10 \pm \sqrt{420}}{2(-16)} = \frac{-10 \pm 20.49}{-32} \]

Step 7 - Solve for the positive root

Calculate the two roots: \[ t_1 = \frac{-10 + 20.49}{-32} = -0.33 \] \[ t_2 = \frac{-10 - 20.49}{-32} = 0.95 \] Since time cannot be negative, t_1 is not valid.

Step 8 - Approximate the solution

Approximate the valid solution to the nearest tenth: \[ t \approx 0.95 \text{ sec} \]

Key Concepts

quadratic formula

The quadratic formula is an essential tool in solving quadratic equations. It's used where the equation is in the standard form \[ ax^2 + bx + c = 0 \]The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here:

• a is the coefficient of \(x^2\)
• b is the coefficient of \(x\)
• c is the constant term

It helps in finding the roots of the quadratic equation. The \(\pm \) symbol means that there will be two solutions: one with addition and one with subtraction. Each solution to the quadratic equation represents an x-value where the curve intersects the x-axis, also known as the roots or zeros of the quadratic equation.

discriminant

The discriminant is part of the quadratic formula and is found under the square root sign: \[ \Delta = b^2 - 4ac \]It provides valuable information about the nature of the roots of the quadratic equation without the need to compute the whole formula.

• If \(\Delta > 0\), there are two distinct real roots.
• If \(\Delta = 0\), there is one real root (the roots are equal).
• If \(\Delta < 0\), the quadratic equation has no real roots; the solutions are complex numbers.

In our ball tossing example, the discriminant was calculated as:\[ 420 \]Since it is greater than 0, we conclude that there are two real roots for the time the ball reaches the ground. However, only the positive root is physically meaningful in this context.

kinematic equations

Kinematic equations describe the motion of objects under constant acceleration. In the context of our problem, the motion of a ball tossed into the air is modeled by one such kinematic equation:\[ h(t) = -\frac{1}{2}gt^2 + vt + h_0 \]where

• h(t) is the height at time t
• g is the acceleration due to gravity (32 \(\text{ft/sec}^2 \) in this case)
• v is the initial velocity (10 \(\text{ft/sec} \) here)
• h_0 is the initial height (5 feet here)

Using this equation, we can derive the height function for our particular problem:\[ h(t) = -16t^2 + 10t + 5 \]We then set the height h(t) to 0 to find when the ball hits the ground, solving the quadratic equation for time, which illustrates how physical principles can be used in conjunction with algebraic techniques.