Problem 23 - Calculus: Graphical, Numerical, Algebraic

Question

Free Fall An object is dropped from the top of a 100 -m tower. Its height above ground after \(t\) sec is \(100-4.9 t^{2}\) m. How fast visit falling 2 sec after it is dropped?

Step-by-Step Solution

Answer

The object would be falling at a speed of 19.6 m/s 2 seconds after it is dropped.

Step 1: Compute the derivative of the height function

The first step is to find the velocity function, which is the derivative of the height function. We can compute this derivative by applying the power rule, which states that the derivative of \(x^n\) with respect to \(x\) is \(n*x^{n-1}\). So if \(h(t) = 100 - 4.9t^{2}\), then its derivative \(v(t)\), which represents the velocity of the object at time \(t\), is given by \(v(t) = -2*4.9*t\). Therefore, \(v(t) = -9.8t\).

Step 2: Plug in the specified time in the velocity function

The next step is to find the velocity of the object 2 seconds after it was dropped. This can be done by substituting \(t = 2\) in the velocity function obtained in the previous step. So, we get \(v(2) = -9.8*2\).

Step 3: Compute the velocity

We can find the velocity of the object by calculating the value obtained in the previous step. So, we find that \(v(2) = -19.6 m/s\).

Key Concepts

Derivative of the Height Function

When studying free fall in calculus, one of the key concepts to understand is the derivative of the height function. The height function describes how high the object is above ground after a certain time since it was dropped. In our example, the height function for an object dropped from a 100-meter tower is given by h(t) = 100 - 4.9t^2, where t is the time in seconds. To determine how fast the object is falling at any given moment, we need to find the velocity function by taking the derivative of the height function.

The derivative represents the rate of change of the function, which, in this context, translates to the velocity or speed of the falling object. In simple terms, by differentiating the height function with respect to time, we are calculating how the height is changing over time — a concept crucial to understanding motion under the influence of gravity.

Velocity Function

Following the concept of the derivative of the height function, let's discuss the velocity function in more detail. The velocity function, denoted as v(t), tells us the speed at which the object is falling at any given time t. It's determined by taking the derivative of the height function h(t).

For the object dropped from the tower, we found that the velocity function is v(t) = -9.8t. The negative sign indicates that the object is moving in the direction opposite to the positive height direction, which makes sense because the object is falling downwards. By calculating v(t), we can determine the falling speed of the object at any instant. For instance, to find the velocity after 2 seconds, we simply substitute t with 2 in the velocity function, yielding v(2) = -9.8 * 2 = -19.6 m/s, which tells us that the object is falling at a speed of 19.6 meters per second after 2 seconds.

Power Rule of Differentiation

The power rule of differentiation is an essential tool for finding the derivative of polynomials, and it's particularly useful in physics problems involving motion like our free fall example. It states that if you have a function f(x) = x^n, where n is any real number, the derivative of f(x) with respect to x is n*x^(n-1).

In our example, the power rule is applied to the term -4.9t^2 in the height function. By using the power rule, we take the exponent 2, multiply it by the coefficient -4.9, and reduce the exponent by 1 to get the derivative -2*4.9t = -9.8t, which is our velocity function. The power rule simplifies the process of differentiation and is a fundamental concept that students will encounter frequently in calculus-related problems involving rates of change.