Object's Descent: Time To Return To Earth

Hey guys! Let's dive into a classic physics problem, shall we? We're going to break down this question about an object launched upwards and figure out when it's going to hit the ground again. It's a fun application of quadratic equations, and trust me, it's not as scary as it sounds. We'll go step-by-step, so even if you're not a math whiz, you'll be able to follow along. The question throws at us an object being launched upwards, and we're given a formula that describes its height at any given time. Our goal? To find out exactly when that object kisses the ground again. Think of it like tossing a ball up in the air; we want to know when it lands back in our hands. Ready? Let's do this!

Understanding the Problem: The Quadratic Equation's Role

Alright, so the core of this problem revolves around understanding what the equation h(t) = -5t² + 20t actually means. The 'h(t)' part represents the height of the object at any given time 't'. The equation itself is a quadratic equation, which is super important because it describes the path of a projectile, like our object. It's a curve, specifically a parabola, that opens downwards. This makes sense because the object goes up, reaches a maximum height, and then comes back down. The coefficients in the equation (-5 and 20 in this case) give us information about the acceleration due to gravity and the initial upward velocity. The key thing to remember is that the object returns to the ground when its height (h(t)) is zero. So, our main task is to find the values of 't' that make h(t) equal to zero. This is where our algebra skills come into play. We're essentially solving a quadratic equation to find the roots, or the points where the parabola intersects the x-axis (in this case, the time axis). It's all about finding those specific moments in time when the object's height is zero, which means it's back on the ground. Think of the equation as a map that tells us exactly where the object is at any moment during its journey.

To really nail this down, imagine you're watching the object in slow motion. At first, it's gaining altitude, but gravity is constantly pulling it down, slowing its ascent. At a certain point, it stops going up and starts coming back down. The time at which it reaches zero height again is what we're after, and that's exactly what our quadratic equation is going to tell us. So, don't sweat the math; just focus on what the equation represents – the object's journey from launch to landing. The equation is our tool, and solving it will give us the answer we need, like finding treasure on a map. Isn't that cool?

Solving for Time: Finding When h(t) = 0

Now, let's get down to the actual solving, shall we? We've established that the object hits the ground when h(t) = 0. So, we'll set our equation equal to zero: -5t² + 20t = 0. This is the heart of the matter, and the way we deal with it is by finding the values of 't' that satisfy this equation. There are a couple of ways to solve this, but we'll use factoring because it's usually the easiest method in cases like this. We can factor out a common term from both parts of the equation. In this case, we can factor out -5t. Doing this, we get -5t(t - 4) = 0. This factored form tells us that the product of -5t and (t - 4) is zero. For this to be true, either -5t = 0 or (t - 4) = 0. Let's solve each of these scenarios separately. First, if -5t = 0, then t = 0. This makes perfect sense! At the moment the object is launched (t = 0 seconds), its height is also zero; it's right on the ground. The second possibility is if (t - 4) = 0, which means t = 4 seconds. This is the answer we are looking for! It tells us that the object returns to the ground after 4 seconds. The object starts at ground level (t=0), goes up, and then comes back down, hitting the ground again at t=4 seconds. So, the solution is not just an abstract number; it's a specific moment in the object's journey. Remember, understanding what the numbers represent is just as crucial as the calculation itself. We used factoring to get to the answer, but knowing why we factored and what the result means is the real win here.

Now, let's pause and appreciate what we've done. We've taken a seemingly complex physics problem and broken it down into manageable steps. By understanding the equation, recognizing what it represents, and applying a simple factoring technique, we found the exact moment when the object returned to Earth. The math wasn't about memorizing formulas; it was about using them to understand a real-world scenario. From there, we got the final time it took the object to hit the ground again, the solution is 4 seconds. Pretty neat, right? The point here is that we can solve problems like this, not by memorizing a bunch of steps, but by really understanding what's going on. This is where we can make a strong connection between physics and math.

Conclusion: The Answer and Why It Matters

So, after all that, we can now confidently say that the object returns to the ground after 4 seconds. The correct answer, based on the options provided, is (C) 4s. That's it, guys! We have successfully tackled the problem. But the real takeaway here isn't just the answer; it's the process. We've learned how to interpret a quadratic equation, connect it to a real-world scenario, and use mathematical tools to solve for a specific value. This kind of problem-solving is applicable far beyond just math class. Think about it: breaking down complex problems, understanding the variables involved, and systematically finding solutions – these skills are valuable in almost any field. The ability to model real-world situations with mathematical equations gives us a powerful tool for understanding and predicting how things work. So, next time you see a problem like this, remember the steps we took. Remember that math is a language, and equations are sentences that describe the world around us. And that's all, folks! Hope you had fun and learned something new today. Until next time, keep exploring and questioning! You can always try to solve other questions and practice more.

Important notes:

• Understanding the Equation: The equation h(t) = -5t² + 20t is a simplified model. In the real world, factors like air resistance might slightly affect the object's trajectory.
• Units: Always pay attention to units. In this case, we're working with meters and seconds, but other problems might use different units.
• Practice: The more you practice these types of problems, the easier they become. Try changing the initial velocity or the coefficient of the t² term and see how the solution changes.

Good luck, and keep up the great work! And hopefully, you'll be able to solve similar questions later on. Have fun, and be curious! That's the best way to learn!