Hey guys! Let's dive into a super cool physics problem involving model rockets! We've got a scenario where a model rocket is launched straight up with an initial velocity of 215 feet per second (ft/s). The rocket's height, which we'll call h, in feet, after t seconds, is described by a neat little equation:
h = 215t - 16t^2
Our mission, should we choose to accept it, is to figure out at what times (t) the rocket will be at a height of 97 feet. This is a classic quadratic equation problem, and we're going to break it down step by step. So, grab your thinking caps, and let's get started!
Setting Up the Equation
Okay, so the first thing we need to do is set up our equation. We know the height (h) we're interested in is 97 feet. So, we're going to substitute 97 for h in our equation:
97 = 215t - 16t^2
Now, this looks a bit messy, right? We want to rearrange it into the standard quadratic equation form, which is:
ax^2 + bx + c = 0
In our case, t is our variable (like x), and we need to get everything on one side of the equation. Let's add 16t^2 to both sides and subtract 215t from both sides to get:
16t^2 - 215t + 97 = 0
Awesome! Now we have a quadratic equation ready to solve. But how do we solve it? There are a few ways we can go about this, and we'll explore the most common methods.
Solving the Quadratic Equation
Method 1: The Quadratic Formula
The quadratic formula is our trusty go-to method for solving any quadratic equation. It might look a little intimidating at first, but trust me, it's a lifesaver. Here's the formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a, b, and c are the coefficients from our quadratic equation (16t^2 - 215t + 97 = 0). So, in our case:
- a = 16
- b = -215
- c = 97
Let's plug these values into the quadratic formula:
t = (-(-215) ± √((-215)^2 - 4 * 16 * 97)) / (2 * 16)
Okay, let's simplify this step by step. First, let's deal with the stuff inside the square root:
(-215)^2 = 46225
4 * 16 * 97 = 6208
So, inside the square root, we have:
46225 - 6208 = 40017
Now, our equation looks like this:
t = (215 ± √40017) / 32
Next, we need to find the square root of 40017. Using a calculator, we get approximately:
√40017 ≈ 200.04
Now we can plug that back into our equation:
t = (215 ± 200.04) / 32
This gives us two possible solutions for t because of the ± sign:
Solution 1:
t = (215 + 200.04) / 32
t = 415.04 / 32
t ≈ 12.97 seconds
Solution 2:
t = (215 - 200.04) / 32
t = 14.96 / 32
t ≈ 0.47 seconds
So, we have two times when the rocket is at 97 feet: approximately 0.47 seconds and 12.97 seconds. This makes sense because the rocket goes up to 97 feet on its way up and then again on its way down.
Method 2: Factoring (If Possible)
Sometimes, we can solve quadratic equations by factoring. Factoring is like reverse multiplication. We try to break down the quadratic expression into two binomials that multiply together to give us the original expression. However, factoring isn't always easy or even possible, especially when we have messy coefficients like in our equation (16t^2 - 215t + 97 = 0).
In this case, factoring would be quite challenging because of the large numbers and the fact that there aren't obvious factors that jump out. So, while factoring is a great technique to have in your toolbox, the quadratic formula is often the more reliable method for equations like this one.
Method 3: Completing the Square
Completing the square is another method for solving quadratic equations. It involves manipulating the equation to create a perfect square trinomial on one side. This method can be a bit more involved, but it's a good technique to understand. However, for this particular equation (16t^2 - 215t + 97 = 0), completing the square would also be quite cumbersome due to the coefficients.
So, for our rocket problem, the quadratic formula is definitely the most straightforward and efficient way to find the solutions.
Interpreting the Results
Okay, so we've crunched the numbers and found two values for t: approximately 0.47 seconds and 12.97 seconds. But what do these numbers actually mean in the context of our rocket launch?
Well, remember that our equation describes the height of the rocket over time. The rocket is launched upwards, reaches a maximum height, and then comes back down. So, the two times we calculated correspond to the rocket passing through the 97-foot mark on its way up and again on its way down.
- t ≈ 0.47 seconds: This is the time when the rocket is at 97 feet on its way up.
- t ≈ 12.97 seconds: This is the time when the rocket is at 97 feet on its way down.
This makes intuitive sense, right? The rocket quickly shoots up, passes 97 feet, continues upwards, reaches its peak, and then falls back down, passing 97 feet again on its descent.
Checking Our Answers
It's always a good idea to check our answers to make sure they make sense and that we haven't made any calculation errors. We can do this by plugging our values of t back into the original equation:
h = 215t - 16t^2
Check for t ≈ 0.47 seconds:
h ≈ 215(0.47) - 16(0.47)^2
h ≈ 100.55 - 16(0.2209)
h ≈ 100.55 - 3.5344
h ≈ 97.0156 feet
This is very close to 97 feet, so our first solution seems good!
Check for t ≈ 12.97 seconds:
h ≈ 215(12.97) - 16(12.97)^2
h ≈ 2788.55 - 16(168.2209)
h ≈ 2788.55 - 2691.5344
h ≈ 97.0156 feet
Again, this is very close to 97 feet, so our second solution also looks good! The slight differences are due to rounding errors in our calculations.
Real-World Considerations
While our mathematical model gives us a good idea of the rocket's trajectory, it's important to remember that real-world situations are often more complex. Our equation doesn't take into account things like air resistance, wind, or changes in the rocket's mass as it burns fuel. These factors could affect the rocket's actual height and the times at which it reaches 97 feet.
However, for a simple model rocket problem, our equation provides a pretty accurate approximation. And, hey, we got to use the quadratic formula, which is always a win!
Conclusion
So, there you have it! We've successfully found the times at which our model rocket reaches a height of 97 feet. By setting up the equation, using the quadratic formula, and interpreting our results, we've solved a classic physics problem. Remember, the key is to break down the problem into smaller, manageable steps, and don't be afraid to use your tools (like the quadratic formula) to get the job done.
Keep exploring, keep questioning, and keep launching those rockets (safely, of course!). Until next time!