Problem 80 - Algebra 1

Question

Use a vertical motion model to find how long it will take for the object to reach the ground. You throw a ball downward with an initial speed of 10 feet per second out of a window to a friend 20 feet below. Your friend does not catch the ball.

Step-by-Step Solution

Answer

It will take approximately 1.264 seconds for the ball to reach the ground.

Step 1: Write down the given values

Given: Initial height (\(h_0\)) is 20 feet below, thus is -20 feet. Initial velocity (\(v\)) is 10 feet/second downward, thus -10 feet/second. The acceleration due to gravity (\(g\)) is 32.2 feet/second^2. Final height (\(h\)) is 0 (ground level). We need to calculate the time (\(t\)).

Step 2: Plug in the values into the equation

Substitute the given values into the equation: 0 = -20 - 10t + 0.5*(32.2)t^2. This simplifies to: 0 = -20 -10t +16.1t^2. Which can be rearranged to: 16.1t^2 -10t - 20 = 0.

Step 3: Apply the Quadratic Formula

Solve the quadratic equation with the Quadratic Formula: \(t = (-b±\sqrt{b^2 - 4ac}) / (2a)\), which gives: \(t = (10±\sqrt{(-10)^2 - 4*16.1*(-20)}) / (2*16.1)\). However, since \(t\) cannot be negative we discard the negative solution, resulting in: \( t = (10 + \sqrt{100 + 1296}) / 32.2 \) .

Step 4: Calculate the Time

Now, calculate the value for \(t\): \( t = (10 + \sqrt{1396}) / 32.2 \). This simplifies further to: \(t = 1.264\) seconds when rounded to three decimal places.

Key Concepts

Quadratic Equation

In vertical motion problems, a quadratic equation often comes into play. Here, the equation takes the form of a parabola describing the object's height relative to time. It appears as:

• \( ax^2 + bx + c = 0 \)

where \(a\), \(b\), and \(c\) are constants, representing the acceleration due to gravity, initial velocity, and initial height, respectively.
This form of equation is solved to find specific variables, such as time (\(t\)), which in turn tells us when the object hits the ground. To solve it, we use methods like factoring, completing the square, or the quadratic formula.
In the example, the quadratic formula was the best choice:

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

This formula simplifies solving when the equation can't be factored easily.

Acceleration Due to Gravity

Gravity is a relentless force pulling objects towards the Earth's center. In most physics problems, this acceleration is considered constant. For the exercise, gravity is given as 32.2 feet per second squared (\(32.2 \text{ ft/s}^2\)).
This value enters the quadratic equation's term related to \(a\), influencing how quickly the object speeds up as it falls. Essentially, it acts as the accelerator in our vertical motion model.
The greater the gravitational acceleration, the faster an object will fall. Ignore air resistance for simplification, though in real-world applications, it might affect the actual result.

Initial Velocity

Initial velocity is where it all begins. It's the speed at which you throw the ball downward. Here, it's 10 feet per second downward, or \(-10 \text{ ft/s}\).
This value influences the object's starting motion alongside gravity, essentially describing its initial momentum.
In our equation, initial velocity contributes to the linear term \(-b\). It either positively or negatively impacts the time-to-ground calculation, depending on the vector of the throw – upward, downward, or horizontal.

Time of Flight

Time of flight represents how long the journey lasts. In our case, it's the duration from when the ball is thrown to when it hits the ground.
Given the calculated quadratic equation \(16.1t^2 - 10t - 20 = 0\), solving using the quadratic formula yielded \(t = 1.264\) seconds.
This calculated time tells us precisely how long the ball remains in the air before hitting the ground. It's a crucial value in predicting motion paths and in physics problems assessing how fast or high things travel.