Problem 42 - Beginning and Intermediate Algebra

Question

An object is launched upward from the ground with an initial velocity of \(200 \mathrm{ft} / \mathrm{sec} .\) The height \(h\) (in feet) of the object after \(t\) sec is given by \(h(t)=-16 t^{2}+200 t\) a) Find the height of the object after 1 sec. b) Find the height of the object after 4 sec. c) When is the object to 400 ft above the ground? d) How long does it take for the object to hit the ground?

Step-by-Step Solution

Answer

a) The height of the object after 1 sec is 184 ft. b) The height of the object after 4 sec is 544 ft. c) The object is 400 ft above the ground at approximately \(t \approx 2\) sec and \(t \approx 10\) sec. d) The object hits the ground after approximately 10 seconds.

Step 1: Recall the given function

We are given the height function, \(h(t)=-16t^2+200t\), which describes the height of an object during its flight in feet as a function of time in seconds.

Step 2: Find the height of the object after 1 sec

To find the height of the object after 1 second, plug in \(t=1\) into the given function: \[h(1)=-16(1)^2+200(1).\] Now, simplify to obtain the height: \[h(1)=-16+200=184.\] So, the height of the object after 1 second is 184 feet.

Step 3: Find the height of the object after 4 sec

Similarly, to find the height of the object after 4 seconds, plug in \(t=4\) into the given function: \[h(4)=-16(4)^2+200(4).\] Now, simplify to obtain the height: \[h(4)=-16(16)+800=-256+800=544.\] So, the height of the object after 4 seconds is 544 feet.

Step 4: Find when the object is 400 ft above the ground

To find when the object reaches 400 ft above the ground, set the height function equal to 400 and solve for t: \[400=-16t^2+200t.\] Since this is a quadratic equation, we can rearrange it by subtracting 400 from both sides: \[0=-16t^2+200t-400.\] Now, we can factor out a 16 from the entire equation to simplify it: \[0=16(-t^2 + 12.5t - 25).\] To find the values of t, we'll have to either factor the quadratic or use the quadratic formula. In this case, it is easier to use the quadratic formula, given by: \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a},\] where \(a=-1\), \(b=12.5\), and \(c=-25\). Substitute these values into the formula and solve for t: \[t=\frac{-12.5\pm\sqrt{(12.5)^2-4(-1)(-25)}}{-2}.\] After calculating, we get two values for t: \(t\approx 2\) and \(t\approx 10\). These represent two times at which the object is 400 ft above the ground: approximately after 2 seconds and after 10 seconds.

Step 5: Determine when the object hits the ground

To find when the object hits the ground, set the height function equal to 0 and solve for t: \[0=-16t^2+200t.\] We've already simplified this equation in step 4 to: \[0=16(-t^2+12.5t-25).\] We can use either the factored equation or the two values we found for t (2 and 10) in step 4. Since the object is launched from the ground and returns to the ground, the two time values represent the times at which the object is at the ground. Therefore, the object hits the ground after approximately 10 seconds.

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object that is thrown or projected into the air, subject to only the acceleration of gravity. This kind of motion in two-dimensional space exhibits a parabolic path under the influence of gravity. In our exercise, the object is launched upwards, making the path symmetrical around its highest point. Understanding this motion involves two fundamental components: vertical and horizontal movements. However, as our equation deals with vertical movement alone, let's focus on that.

• Initial Velocity: This is the speed at which the object is projected upward. In our problem, it's given as 200 ft/sec.
• Acceleration Due to Gravity: Gravity is a constant force pulling the object downward at 32 ft/sec², though in our formula it's incorporated as -16t² due to the conversion factor.

From the moment the object is launched, it will ascend to a certain point before gravity pulls it back down again. Studying this sort of motion is very useful in real-world scenarios, from sports to engineering. For instance, understanding projectile motion can help optimize the angles in shots to achieve maximum distance.

Solving Quadratic Equations

When you're faced with a quadratic equation, you often want to find out at what points the parabola will intersect the horizontal axis, providing valuable information like maximum height or the time it takes to reach a certain altitude or return to the ground. A quadratic equation is generally given in the form:\[ax^2 + bx + c = 0\]where \(a\), \(b\), and \(c\) are constants. The solution of this equation can be found using several methods:

• Factoring: This involves expressing the quadratic as a product of two simpler expressions.
• Quadratic Formula: It provides a straightforward solution using the formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
• Completing the square: A method that involves restructuring the quadratic equation so it's easy to solve for the variable.

In our case, to find when the object reaches 400 ft or returns to the ground, we set the height equation equal to those values and solved for \(t\) using the quadratic formula. This approach is especially convenient for equations that do not factor easily. Quadratic equations are a crucial aspect of projectile motion problems, providing insights into the object's behavior over time.

Applications of Quadratic Functions

Quadratic functions are rich in applications, offering insight into a wide array of real-world scenarios. Anything from physics to economics can be understood better through them. In our exercise, the quadratic function describes the flight of a projectile, highlighting its practical utility. Here are some typical applications:

• **Projectile Motion:** As seen, they help model the trajectory of objects in flight, predicting positions at specific times.
• Sports:** Quadratic functions are used in sports to model arcs, such as the ball's trajectory in basketball or golf.
• Business:** In economics, these functions can represent profit, cost, or revenue models over time to determine optimal production strategies.

Quadratic functions don't just show when something reaches a certain height or speed, but help to maximize or minimize outcomes. For instance, businesses use them to find the most efficient routes or schedules. In engineering, these functions help model structures, ensuring they remain within safe parameters. Overall, learning about quadratic functions not only improves problem-solving skills but also enables us to predict and analyze significant real-life scenarios efficiently.