Understanding H(3.2) In Projectile Motion The Height Of A Rock

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In the realm of physics and mathematics, understanding the motion of objects is a fundamental concept. Projectile motion, in particular, provides a fascinating framework for analyzing the trajectory of objects launched into the air. This article delves into a specific scenario involving the height of a rock propelled by a slingshot, described by the function $h(t) = -16t^2 + 28t + 500$. Our primary focus is to interpret the meaning of $h(3.2)$ within this context, exploring the implications of this mathematical expression in real-world terms.

Let's begin by dissecting the given function, $h(t) = -16t^2 + 28t + 500$. This quadratic equation models the height of the rock, denoted as $h(t)$, at any given time $t$ seconds after it is launched. The coefficients in this equation hold significant physical meanings:

  • -16: This coefficient represents half of the acceleration due to gravity (approximately -32 feet per second squared). The negative sign indicates that gravity acts downwards, causing the rock's upward velocity to decrease over time.
  • 28: This coefficient corresponds to the initial upward velocity of the rock when it is launched from the slingshot. It signifies how fast the rock is initially propelled into the air.
  • 500: This constant term represents the initial height of the rock at time $t = 0$. It indicates the height from which the rock is launched, which could be the height of the person holding the slingshot or the elevation of the ground.

By plugging in different values of $t$ into the function, we can determine the height of the rock at various points in its trajectory. For instance, $h(1)$ would give us the height of the rock 1 second after it is launched, and so on. Understanding the components of this function is crucial for interpreting the meaning of $h(3.2)$.

The core question we aim to address is: What does $h(3.2)$ represent in the context of the rock's trajectory? To answer this, we simply substitute $t = 3.2$ into the height function $h(t)$:

h(3.2)=−16(3.2)2+28(3.2)+500h(3.2) = -16(3.2)^2 + 28(3.2) + 500

Calculating this value will give us the height of the rock at $t = 3.2$ seconds after it is propelled from the slingshot. The result, a numerical value in feet, signifies the vertical distance of the rock from the ground (or the reference point of 500 feet) at that specific moment in time. In essence, $h(3.2)$ provides a snapshot of the rock's position in its flight path exactly 3.2 seconds after its launch. It doesn't tell us anything about the rock's position before or after this specific time; it's a precise measurement of its height at that instant. Therefore, option A, which speaks of the time before the rock reaches the ground, is misleading. $h(3.2)$ represents a particular point in the rock's journey, not a duration or a prediction of the future. Understanding this distinction is key to interpreting projectile motion problems accurately. The value of $h(3.2)$ will be a single number, representing the rock's altitude at that precise moment.

To fully grasp the meaning of $h(3.2)$, let's calculate its numerical value. This will provide a concrete understanding of the rock's height at 3.2 seconds. We'll break down the calculation step-by-step for clarity:

  1. Square the time: First, we calculate $3.2^2$, which equals 10.24.
  2. Multiply by -16: Next, we multiply the result by -16: $-16 * 10.24 = -163.84$. This term represents the effect of gravity on the rock's height.
  3. Multiply 28 by 3.2: Now, we calculate $28 * 3.2 = 89.6$. This term represents the upward velocity component of the rock's motion.
  4. Add the terms: Finally, we add all the terms together: $-163.84 + 89.6 + 500 = 425.76$.

Therefore, $h(3.2) = 425.76$. This means that 3.2 seconds after the rock is propelled from the slingshot, its height is 425.76 feet. This calculation provides a specific data point in the rock's trajectory, giving us a clear picture of its position at that moment. It's important to note that this is just one point in the rock's flight path. To fully understand the rock's motion, we would need to consider its height at various other times as well.

The value of $h(3.2) = 425.76$ feet gives us a specific point on the rock's trajectory. To visualize this, imagine the rock being launched upwards, reaching a peak height, and then descending back towards the ground. At 3.2 seconds after launch, the rock is at a height of 425.76 feet. This likely occurs after the rock has reached its maximum height and is on its descent, since the initial height was already 500 feet. Understanding the context of projectile motion helps us interpret this value. The parabolic path of the rock is symmetrical, but the presence of the initial height of 500 feet skews the symmetry slightly. Therefore, knowing the height at one point in time helps us infer the rock's position relative to its peak and its landing point. Furthermore, comparing $h(3.2)$ with heights at other times would provide a more complete picture of the rock's motion. For example, calculating $h(1)$ or $h(2)$ would reveal how the rock's height changes over time, allowing us to understand its velocity and acceleration at different stages of its flight.

When dealing with projectile motion functions like $h(t)$, several common misinterpretations can arise. One frequent error is assuming that $h(3.2)$ represents the time it takes for the rock to reach the ground. As we've established, $h(3.2)$ specifically represents the height of the rock at 3.2 seconds, not the total flight time. To find the time when the rock hits the ground, we would need to solve the equation $h(t) = 0$, which is a different problem altogether. Another misconception is thinking that $h(3.2)$ gives information about the rock's horizontal distance. The function $h(t)$ solely describes the vertical position (height) of the rock; it doesn't account for any horizontal movement. To analyze the horizontal motion, we would need a separate function that describes the rock's horizontal position as a function of time. Additionally, it's crucial to remember that these functions are simplified models of reality. They often neglect factors such as air resistance, which can significantly affect the actual trajectory of the rock. Therefore, while $h(t)$ provides a valuable approximation, it's essential to recognize its limitations. Avoiding these misinterpretations is key to accurately solving projectile motion problems and applying the concepts to real-world scenarios.

In conclusion, $h(3.2)$ represents the height of the rock 3.2 seconds after it is propelled from the slingshot. This single value provides a snapshot of the rock's vertical position at a specific moment in time. By calculating $h(3.2)$, we determined that the rock is at a height of 425.76 feet at 3.2 seconds after launch. This understanding is crucial in analyzing projectile motion, as it allows us to track the object's trajectory and predict its position at various points in its flight. However, it's important to avoid common misinterpretations and remember that $h(3.2)$ is just one piece of the puzzle. To fully comprehend the rock's motion, we need to consider the height function as a whole and analyze the rock's position at multiple times. By carefully interpreting the mathematical representation and connecting it to the physical context, we can gain a deeper understanding of the principles governing projectile motion.