Quadratic Model: Ball Drop Height Problem
Hey everyone, let's dive into a cool math problem today involving a quadratic model. We're going to figure out when a dropped ball reaches a certain height. This kind of problem pops up a lot in physics and engineering, so understanding how to tackle it is super useful, guys. The quadratic model we're working with is f(x) = -5x² + 200. This equation tells us the approximate height, in meters, of a ball x seconds after it's been dropped. Our main mission is to find out after how many seconds the ball will be 50 meters from the ground. We've got some multiple-choice options to help us narrow it down: A. 2.45 seconds, B. 3.16 seconds, C. 5.48 seconds, and D. 7.07 seconds. Ready to crunch some numbers and solve this?
Understanding the Quadratic Model and the Problem
So, what's the deal with this f(x) = -5x² + 200 equation, you ask? Well, it's a quadratic model, and it's designed to describe the path of an object under the influence of gravity, neglecting air resistance. In this specific scenario, f(x) represents the ball's height in meters, and x is the time in seconds after it's dropped. The '-5x²' part is what makes it quadratic, and it's responsible for that parabolic trajectory. The '+ 200' is the initial height from which the ball was dropped. Think of it like this: at time x=0 (the moment it's dropped), the height is f(0) = -5(0)² + 200 = 200 meters. So, the ball starts its journey from a height of 200 meters. Our goal is to find the specific time, x, when the height, f(x), is exactly 50 meters. This means we need to set our equation equal to 50 and solve for x. It's like asking, "At what point in its descent does the ball hit the 50-meter mark?" This is a classic application of quadratic equations, and it shows how math can model real-world phenomena. We're not just doing abstract calculations here; we're simulating a physical event. The fact that it's a negative coefficient for the x² term (-5) indicates that the parabola opens downwards, which makes sense for an object falling under gravity – its height decreases over time. The value '-5' is related to the acceleration due to gravity, but simplified for this problem. So, to solve this, we'll be plugging in 50 for f(x) and then isolating x. Let's get our algebra hats on!
Setting Up the Equation to Solve
Alright, team, we've got our model: f(x) = -5x² + 200. We know that f(x) is the height, and we want to find the time x when the height is 50 meters. So, we simply substitute 50 for f(x) in our equation. This gives us: 50 = -5x² + 200. Now, our objective is to isolate x to find out how many seconds it takes for the ball to reach this height. This is where the fun of algebraic manipulation comes in! We need to get x² by itself on one side of the equation. First, let's move the constant term (200) to the other side. To do this, we subtract 200 from both sides of the equation:
50 - 200 = -5x² + 200 - 200
This simplifies to:
-150 = -5x²
See how we're getting closer? Now, x² is being multiplied by -5. To get x² alone, we need to divide both sides of the equation by -5:
-150 / -5 = -5x² / -5
Performing the division, we get:
30 = x²
So, we've found that x² = 30. This is a crucial step because it tells us the square of the time is 30. The next logical step is to find the value of x itself. Remember, x represents time, and time is always a non-negative value (we're looking at seconds after the ball is dropped, not before). Therefore, we need to take the square root of both sides of the equation to solve for x:
√x² = √30
This gives us:
x = √30
Now, we just need to calculate the square root of 30. This is where we'll likely need a calculator, or we can estimate if we're feeling brave. The exact value of √30 is what we're looking for to compare with our answer choices. Keep in mind that in the real world, we usually round these values to a certain number of decimal places for practical purposes. So, let's go find that approximate value!
Calculating the Time and Choosing the Best Answer
Okay guys, we've reached the point where x² = 30, and we need to find x by taking the square root of 30. Using a calculator, we find that the square root of 30 is approximately 5.477225575... seconds. Now, let's look at our answer choices:
- A. 2.45 seconds
- B. 3.16 seconds
- C. 5.48 seconds
- D. 7.07 seconds
Comparing our calculated value (~5.477 seconds) to the options, we can see that option C. 5.48 seconds is the closest approximation. The difference is just due to rounding. When we calculate √30, it's about 5.477. Rounding this to two decimal places gives us 5.48.
Let's double-check our work by plugging 5.48 back into the original equation to see if we get close to 50 meters.
f(5.48) = -5 * (5.48)² + 200
First, square 5.48:
5.48² ≈ 30.0304
Now, multiply by -5:
-5 * 30.0304 ≈ -150.152
Finally, add 200:
-150.152 + 200 ≈ 49.848
This result, 49.848 meters, is very close to our target of 50 meters. The slight difference is due to the rounding of the time value. If we had used the more precise value of √30, we would get exactly 50. This confirms that our calculation and the choice of 5.48 seconds are correct.
It's also worth noting why the other options are incorrect. If we take option A (2.45 seconds), f(2.45) = -5*(2.45)² + 200 = -5*6.0025 + 200 = -30.0125 + 200 = 169.9875 meters. Way too high! For option B (3.16 seconds), f(3.16) = -5*(3.16)² + 200 = -5*9.9856 + 200 = -49.928 + 200 = 150.072 meters. Still too high. For option D (7.07 seconds), f(7.07) = -5*(7.07)² + 200 = -5*49.9849 + 200 = -249.9245 + 200 = -49.9245 meters. This is nonsensical as height can't be negative in this context (it implies the ball would have hit the ground and gone below it, which isn't how the model works past hitting ground level). So, 5.48 seconds is definitely our winner!
Conclusion: Mastering Quadratic Models in Real-World Problems
So there you have it, folks! We successfully tackled a problem involving a quadratic model that describes the height of a dropped ball over time. We started with the given equation, f(x) = -5x² + 200, where f(x) is the height in meters and x is the time in seconds. Our goal was to find the time x when the ball is 50 meters from the ground. By setting f(x) = 50, we transformed the problem into an algebraic equation: 50 = -5x² + 200. Through a series of steps – subtracting 200 from both sides to get -150 = -5x², and then dividing by -5 to get x² = 30 – we isolated x². The final step involved taking the square root of both sides to solve for x, giving us x = √30. Calculating this value, we found it to be approximately 5.477 seconds. When compared to the multiple-choice options, 5.48 seconds was the closest and therefore the correct answer. We even did a quick check by plugging 5.48 back into the original equation, which yielded a height very close to 50 meters, confirming our solution.
This problem highlights the power of quadratic models in representing real-world physical situations. Whether it's projectile motion, optimization problems, or understanding the behavior of systems influenced by squared variables, quadratic equations are everywhere. Understanding how to set up, manipulate, and solve these equations is a fundamental skill in mathematics and science. Remember, always keep an eye on the context of the problem – in this case, time must be positive. And don't be afraid of using a calculator for those square roots! Practice makes perfect, so try working through similar problems. The more you practice, the more comfortable you'll become with these concepts, and you'll be solving physics and math puzzles like a pro. Keep exploring, keep learning, and remember that math is a tool to understand the world around us!