Problem 37 - Brief Applied Calculus

Question

Impact Velocity If a steel ball is tossed from the top of the Burj Khalifa in Dubai, the tallest building in the world, its height above the ground \(t\) seconds after it is dropped will be \(s(t)=2717-16 t^{2}\) feet (neglecting air resistance). a. How long will it take to reach the ground? [Hint: Find when the height equals zero.] b. Use your answer to part (a) to find the velocity with which it will strike the ground. (This is called the impact velocity.) c. Find the acceleration at any time \(t\). (This number is called the acceleration due to gravity.)

Step-by-Step Solution

Answer

a. 13.05 seconds; b. -417.6 ft/s; c. -32 ft/s².

Step 1: Find Time to Reach Ground

To determine how long it will take for the ball to reach the ground, we set the height function equal to zero. \[ 2717 - 16t^2 = 0\]Solve for \(t\): \[ 16t^2 = 2717\]\[ t^2 = \frac{2717}{16}\]\[ t = \sqrt{\frac{2717}{16}}\]Calculate the value:\[ t \approx 13.05 \text{ seconds}\]

Step 2: Calculate Impact Velocity

The velocity function is the derivative of the height function, \(s(t)\). So, differentiate:\[ v(t) = \frac{d}{dt}(2717 - 16t^2) = -32t\]Use the time found in Step 1 to find the impact velocity:\[ v(13.05) = -32 \times 13.05 \]\[ v(13.05) \approx -417.6 \text{ feet per second}\]

Step 3: Determine Acceleration Due to Gravity

The acceleration is the derivative of the velocity function, \(v(t) = -32t\). Therefore, differentiate:\[ a(t) = \frac{d}{dt}(-32t) = -32 \]The acceleration is constant and called the acceleration due to gravity, which is \(-32\) feet per second².

Key Concepts

Impact Velocity

When an object falls from a height, like a steel ball dropped from the Burj Khalifa, its speed just before hitting the ground is known as the impact velocity. To calculate this, we use the derivative of the object's height function since velocity is a change of height over time.

In our problem, the height of the ball is given by the function \(s(t) = 2717 - 16t^2\). The derivative of the height function gives us the velocity function:

• Velocity, \(v(t) = \frac{d}{dt}(2717 - 16t^2) = -32t\)

Using the time to reach the ground (about 13.05 seconds), the impact velocity is:

• Impact Velocity, \(v(13.05) = -32 \times 13.05 \approx -417.6 \text{ feet per second}\)

This negative sign indicates the direction of the velocity, which is downward towards the ground. Thus, when an object impacts the ground, it does so with substantial speed, influenced by factors like the height from which it falls.

Acceleration Due to Gravity

The acceleration due to gravity is a measure of how quickly a falling object's velocity increases as it falls to Earth. On our planet, this value is a constant that every freely falling object experiences, ignoring air resistance.

In the calculation, we find acceleration by taking the derivative of the velocity function. The velocity function, derived from the height function, is \(v(t) = -32t\). Calculating its derivative gives us:

• Acceleration, \(a(t) = \frac{d}{dt}(-32t) = -32\)

Thus, the acceleration due to gravity in this problem is \(-32\) feet per second squared (ft/s²). This number indicates that every second, the object's velocity increases by 32 feet per second in a downward direction. Acceleration due to gravity is essential in predicting how objects move when falling freely and is the same for any object, no matter its mass.

Derivative in Calculus

Calculus is a powerful mathematical tool that lets us understand changes, like how an object's position changes over time. In our context, we use derivatives to find both velocity and acceleration from a position (or height) function.

A derivative represents the rate of change. When finding the velocity function from the height, we take the derivative of the height function \(s(t) = 2717 - 16t^2\) to get:

• Velocity, \(v(t) = \frac{d}{dt}(s(t)) = -32t\)

To find acceleration, we further differentiate the velocity function:

• Acceleration, \(a(t) = \frac{d}{dt}(v(t)) = -32\)

This demonstrates the chain of derivation: from position to velocity, and from velocity to acceleration. Understanding derivatives is crucial as they provide insight into natural laws like gravity and motion.