Calculating Rock Height After 3 Seconds A Free Fall Physics Problem

In the realm of physics, understanding the motion of objects under the influence of gravity is crucial. This article delves into the scenario of a rock falling off a cliff, exploring the concepts of free fall and how to calculate the height of the rock at a specific time. We will use the provided functions to determine the rock's height above the ground after 3 seconds. This problem combines initial conditions and the derivative of a function to find a specific value, illustrating a fundamental application of calculus in physics.

The provided information gives us two key pieces of data: the initial height of the rock and the rate at which its height changes over time. The function f(t) represents the height of the rock in meters above the ground after t seconds. The initial condition, f(0) = 70, tells us that the rock starts at a height of 70 meters. The derivative of the function, f'(t) = -9.8t, represents the rock's instantaneous vertical velocity at time t. The negative sign indicates that the velocity is directed downwards, as expected in free fall. The value 9.8 m/s² is the approximate acceleration due to gravity near the Earth's surface. To find the height of the rock at any time t, we need to integrate the velocity function and use the initial condition to find the constant of integration. This process will give us the position function f(t), which we can then evaluate at t = 3 seconds. This example demonstrates how calculus can be used to describe and predict motion in a physical context. Understanding the relationship between position, velocity, and acceleration is fundamental in physics, and this problem provides a clear application of these concepts.

Let's restate the problem clearly. We have a rock dropped from a cliff. The function f(t) describes the rock's height above the ground in meters at time t seconds. We know two things: the initial height f(0) = 70 meters and the rate of change of height (the vertical velocity) f'(t) = -9.8t meters per second. Our goal is to determine the height of the rock after 3 seconds, which means finding the value of f(3). This involves using the given velocity function (the derivative) to find the position function f(t) and then substituting t = 3 into the resulting function. This kind of problem is a classic application of calculus in physics, involving the concepts of initial conditions, derivatives, and integrals. By solving it, we demonstrate how we can use mathematical tools to model and predict the motion of objects under the influence of gravity. Understanding this process allows us to extend the concepts to more complex scenarios, such as objects thrown at an angle or moving through air resistance. The clarity of the problem statement is crucial because it guides the solution process. Knowing exactly what we need to find and what information we have available is the first step in any problem-solving exercise.

To find the height function f(t), we need to perform the integration of the velocity function, f'(t) = -9.8t. Integration is the reverse process of differentiation, and it allows us to find the original function given its rate of change. The integral of f'(t) with respect to t will give us f(t) plus a constant of integration, usually denoted by C. Mathematically, we write this as ∫ f'(t) dt = f(t) + C. Applying this to our problem, we have ∫(-9.8t) dt. To integrate -9.8t, we use the power rule of integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C. In our case, n = 1, so the integral becomes -9.8 * (t^(1+1))/(1+1) + C, which simplifies to -9.8 * (t^2/2) + C, or -4.9t^2 + C. Thus, f(t) = -4.9t^2 + C. The constant of integration C is crucial because it represents the initial height of the rock, which we know is 70 meters. To find C, we use the initial condition f(0) = 70. Substituting t = 0 into f(t) = -4.9t^2 + C, we get f(0) = -4.9(0)^2 + C = C. Since f(0) = 70, we have C = 70. Therefore, the height function is f(t) = -4.9t^2 + 70. This equation tells us the height of the rock at any time t. Now, to find the height after 3 seconds, we simply substitute t = 3 into the equation. This integration method is a fundamental technique in calculus, and it's widely used in physics to solve problems involving motion and other dynamic systems.

Let's break down the integration process into smaller, more manageable steps. First, we start with the velocity function, f'(t) = -9.8t. This function represents the instantaneous rate of change of the rock's height. To find the height function, f(t), we need to perform the indefinite integral of f'(t) with respect to t. The indefinite integral is represented as ∫f'(t) dt. In our case, this is ∫(-9.8t) dt. The constant -9.8 is a coefficient, so we can take it out of the integral: -9.8∫t dt. Now, we need to integrate t with respect to t. Remember, t can be thought of as t^1. Using the power rule of integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, we get ∫t^1 dt = (t^(1+1))/(1+1) + C = t^2/2 + C. Multiplying this by the coefficient -9.8, we get -9.8 * (t^2/2) + C = -4.9t^2 + C. Thus, the general solution for f(t) is f(t) = -4.9t^2 + C, where C is the constant of integration. This constant is essential because it represents the initial condition, which in this problem, is the initial height of the rock. The step-by-step process ensures that each operation is clearly understood, reducing the chances of error and making the process more accessible.

The constant of integration, often denoted as C, plays a crucial role in finding the specific solution to our problem. After integrating the velocity function f'(t) = -9.8t, we obtained the general solution f(t) = -4.9t^2 + C. However, this is a family of functions, each differing by the value of C. To find the particular function that describes the rock's height, we need to determine the value of C. This is where the initial condition comes into play. We are given that f(0) = 70, which means that at time t = 0 seconds, the height of the rock is 70 meters. We can use this information to solve for C. Substitute t = 0 into the general solution: f(0) = -4.9(0)^2 + C. This simplifies to f(0) = 0 + C, so f(0) = C. Since we know that f(0) = 70, we can conclude that C = 70. This means the specific height function that describes the rock's motion is f(t) = -4.9t^2 + 70. The constant of integration is not just a mathematical artifact; it has a physical meaning. In this context, it represents the initial height of the rock, the starting point of its fall. Without considering this constant, we would only have a general formula for the rock's height, not the specific trajectory of this particular rock falling from this particular cliff. Identifying and determining the constant of integration is a crucial step in solving differential equations and problems involving initial conditions.

An alternative approach involves using definite integrals to calculate the change in height over a specific time interval. Instead of finding the general function f(t) first, we can directly calculate the change in height between time t = 0 and t = 3 seconds. The change in height is given by the definite integral of the velocity function f'(t) from 0 to 3: ∫[0,3] f'(t) dt. This integral represents the net displacement of the rock during this time interval. We know that f'(t) = -9.8t, so we need to evaluate ∫[0,3] (-9.8t) dt. The definite integral is calculated by first finding the indefinite integral and then evaluating it at the upper and lower limits of integration and subtracting the results. As we found earlier, the indefinite integral of -9.8t is -4.9t^2 + C. However, when evaluating a definite integral, the constant of integration C cancels out, so we don't need to worry about it. We evaluate -4.9t^2 at t = 3 and t = 0: (-4.9)(3^2) - (-4.9)(0^2) = -4.9(9) - 0 = -44.1 meters. This tells us that the rock's height decreased by 44.1 meters between t = 0 and t = 3 seconds. Since the initial height was 70 meters, the height after 3 seconds is 70 - 44.1 = 25.9 meters. This method offers a direct way to calculate the change in height without explicitly finding the height function f(t). It relies on the fundamental theorem of calculus, which connects the derivative and the integral of a function. Using definite integrals can be particularly useful when we are only interested in the change in a quantity over a specific interval, rather than the quantity's value at all times.

Let's outline the steps to calculate the definite integral. We want to evaluate ∫[0,3] (-9.8t) dt, which represents the change in the rock's height from t = 0 to t = 3 seconds. The first step is to find the indefinite integral of -9.8t. As we discussed earlier, the indefinite integral of -9.8t is -4.9t^2 + C*. However, in definite integrals, the constant of integration C will cancel out when we evaluate the limits, so we can omit it for this step. Now, we need to evaluate the indefinite integral at the upper and lower limits of integration. The upper limit is 3, and the lower limit is 0. We substitute these values into -4.9t^2: At t = 3, we have -4.9(3^2) = -4.9(9) = -44.1. At t = 0, we have -4.9(0^2) = 0. Next, we subtract the value at the lower limit from the value at the upper limit: -44.1 - 0 = -44.1. Therefore, the definite integral ∫[0,3] (-9.8t) dt is equal to -44.1 meters. This value represents the net change in the rock's height during the time interval from 0 to 3 seconds. The negative sign indicates that the rock's height has decreased, which is expected since it's falling. The definite integral provides a concise way to calculate the accumulated change of a function over an interval, and it's a fundamental tool in calculus with applications in various fields, including physics, engineering, and economics.

Interpreting the result of the definite integral is crucial for understanding its physical significance. We calculated that ∫[0,3] (-9.8t) dt = -44.1 meters. This value represents the net change in the rock's vertical position between time t = 0 seconds and t = 3 seconds. The negative sign is significant. In this context, it indicates that the change in height is a decrease, meaning the rock has moved downwards. The magnitude of the value, 44.1 meters, tells us how much the rock's height has decreased during this time interval. So, from t = 0 to t = 3 seconds, the rock fell a total of 44.1 meters. This is a displacement, a measure of how much the rock's position has changed, not necessarily the total distance it has traveled (although in this case, they are the same since the rock is moving in one direction). If we wanted to find the rock's height at t = 3 seconds, we would need to consider its initial height. Since the initial height is given as 70 meters (f(0) = 70), we subtract the change in height from the initial height: 70 meters - 44.1 meters = 25.9 meters. This means that after 3 seconds, the rock is 25.9 meters above the ground. Interpreting the result of a definite integral in its physical context is essential for translating a mathematical calculation into a meaningful understanding of the situation being modeled. In this case, we've used the definite integral to quantify the rock's fall and determine its position at a specific time.

Now that we have both the height function and the change in height, we can determine the height of the rock after 3 seconds. Using Method 1, we found the height function f(t) = -4.9t^2 + 70. To find the height at t = 3 seconds, we substitute t = 3 into the function: f(3) = -4.9(3^2) + 70 = -4.9(9) + 70 = -44.1 + 70 = 25.9 meters. Using Method 2, we calculated the change in height between t = 0 and t = 3 seconds as -44.1 meters. Since the initial height was 70 meters, the height after 3 seconds is 70 - 44.1 = 25.9 meters. Both methods yield the same answer: the rock is 25.9 meters above the ground after 3 seconds. This consistency reinforces the validity of our calculations and demonstrates how different approaches in calculus can lead to the same result. The final answer provides a concrete value for the rock's position at a specific time, completing the solution to the problem. It's essential to state the answer clearly, along with the appropriate units (meters in this case), to ensure a complete and understandable solution. This problem demonstrates the power of calculus in modeling physical phenomena and predicting the behavior of objects in motion.

In conclusion, by applying the principles of calculus, we successfully determined the height of a rock dropped from a cliff after 3 seconds. We utilized two methods: integrating the velocity function to find the position function and using definite integrals to directly calculate the change in height. Both approaches led to the same answer: the rock is 25.9 meters above the ground after 3 seconds. This problem highlights the practical application of calculus in physics, particularly in understanding motion under the influence of gravity. By understanding concepts like derivatives, integrals, and initial conditions, we can accurately model and predict the behavior of objects in various physical scenarios. The ability to solve such problems is crucial in fields like engineering, physics, and other scientific disciplines. Furthermore, this exercise demonstrates the importance of understanding the relationship between mathematical concepts and their physical interpretations. The negative sign in the change in height, the constant of integration representing initial conditions, and the final answer all have specific meanings in the context of the problem. The problem also demonstrates the flexibility of calculus, offering multiple approaches to arrive at the same solution, thus reinforcing the understanding of the underlying concepts. This understanding allows us to tackle more complex problems in the future, further demonstrating the power and versatility of calculus as a problem-solving tool.

Free fall, gravity, integration, differentiation, initial conditions, velocity, height, position, displacement, definite integral, indefinite integral, constant of integration, calculus, physics, motion, mathematical modeling.