Problem 56 - Essential Calculus

Question

Use \(a(t)=-9.8\) meters per second per second as the acceleration due to gravity. The Grand Canyon is 1600 meters deep at its deepest point. A rock is dropped from the rim above this point. Express the height of the rock as a function of the time \(t\) in seconds. How long will it take the rock to hit the canyon floor?

Step-by-Step Solution

Answer

The height of the rock as a function of time can be expressed as \(h = 1600 - 4.9t^2\). It will take approximately 18.1 seconds for the rock to hit the canyon floor.

Step 1: Identify given values

In this scenario, the key given values are acceleration due to gravity (which is negative because it pulls objects downwards), \(a(t) = -9.8 m/s^2\), and the height from which rock is being dropped, \(h = 1600 m\).

Step 2: Formulate the height function

The formula for height in free fall, where initial velocity is 0 (since the rock is being dropped and not thrown) is given by \(h = h_0 + v_0t - 0.5at^2\). Here, \(h_0\) is the initial height from which the rock is dropped, \(v_0\) is the initial velocity of the rock, \(a\) is the acceleration due to gravity, and \(t\) is the time. Plugging in the given values we get, \(h = 1600 - 0.5*(-9.8)t^2\), which simplifies to \(h = 1600 - 4.9t^2\).

Step 3: Determine the time it takes for the rock to hit the ground

We set \(h=0\) and solve for \(t\) in the equation \(0 = 1600 - 4.9t^2\) which gives us \(t = \sqrt{1600 / 4.9}\). Using a calculator, this gives approximately \(t = 18.1\) seconds.

Key Concepts

Acceleration Due to Gravity

One of the fundamental forces experienced by objects on Earth is the force of gravity, which gives rise to the phenomenon known as 'acceleration due to gravity.' Simply put, it's a measure of how quickly an object speeds up as it falls freely towards the Earth's surface. It's often denoted by the symbol 'g' and has an average value of approximately -9.8 meters per second squared \(m/s^2\) at Earth's surface.

Why is it negative? The negative sign indicates that gravity pulls the object downward, towards the center of the Earth, which is the opposite direction of positive height or position above the ground. This concept is critical in analyzing the motion of freely falling objects, as it is a constant value that affects the speed and overall flight time of any object in free fall. Understanding how to apply this acceleration in kinematic equations allows us to predict the behavior of objects dropped or thrown in an environment where air resistance is negligible.

Kinematic Equations

Kinematic equations describe the motion of objects using variables such as displacement, initial velocity, final velocity, acceleration, and time. These equations are a staple in physics, especially when dealing with free fall scenarios, as they enable us to calculate various parameters of an object's motion.

In a free fall situation, the initial velocity \(v_0\) is often zero if an object is simply dropped, rather than thrown. The general form of the kinematic equation for displacement is \( h = h_0 + v_0t - 0.5at^2 \), where \( h_0 \) is the initial height, \( v_0 \) is the initial velocity, 'a' is the acceleration, and 't' is the time elapsed. When dealing with free fall with the gravity factor, \(v_0\) is zero, and \( 'a' \) takes the value of the acceleration due to gravity. With these equations, we can determine how high an object will go, how long it will stay in the air, or how long it will take for it to hit the ground from a given height.

Quadratic Functions

Quadratic functions are mathematical expressions of the form \(f(x) = ax^2 + bx + c\) where 'a', 'b', and 'c' are constants and the highest power of 'x' is 2. In the context of free fall, the height of an object at any given time can be represented as a quadratic function of time.

For instance, using the kinematic equation for free fall \(h(t) = h_0 - 0.5gt^2\), the variable 't' is squared, making it a quadratic function of time, where \(h_0\) is effectively \(c\), the constant term, \(g\) corresponds to \(a\), and \(b\) would be zero since there's no term involving just 't'. Quadratic functions are essential for calculating the motion parameters in physics because they enable us to model the path of projectiles and objects in free fall. When solving for time in these equations, we often end up with square root calculations, as seen in the given exercise, to find out how long it takes for an object to reach the ground or the peak of its trajectory.