Problem 13 - Thomas Calculus

Question

Free fall from the Tower of Pisa Had Galileo dropped a cannonball from the Tower of Pisa, 179 ft above the ground, the ball's height above ground \(t\) sec into the fall would have been \(s=179-16 t^{2} .\) a. What would have been the ball's velocity, speed, and acceleration at time \(t ?\) b. About how long would it have taken the ball to hit the ground? c. What would have been the ball's velocity at the moment of impact?

Step-by-Step Solution

Answer

Velocity is \(-32t\), acceleration is \(-32\). Hits the ground in \(3.34\) seconds at \(-106.88\) ft/s.

Step 1: Find the Velocity Function

The velocity is given by the derivative of the height function with respect to time. The height function is \(s = 179 - 16t^2\). Taking the derivative with respect to \(t\), we find the velocity function: \(v(t) = \frac{ds}{dt} = -32t\).

Step 2: Determine the Speed

The speed is the absolute value of the velocity. Therefore, from our velocity function \(v(t) = -32t\), the speed at time \(t\) is \(|v(t)| = 32t\).

Step 3: Identify the Acceleration

Acceleration is the derivative of velocity with respect to time. Since \(v(t) = -32t\), the derivative is \(a(t) = \frac{dv}{dt} = -32\). This is a constant acceleration due to gravity.

Step 4: Find Time to Hit the Ground

The ball hits the ground when its height \(s = 0\). Setting the height equation to zero: \(179 - 16t^2 = 0\). Solving for \(t\), we get \(16t^2 = 179\) and \(t^2 = \frac{179}{16}\). Thus, \(t = \sqrt{\frac{179}{16}} \approx 3.34\) seconds.

Step 5: Determine Velocity at Impact

Using the velocity equation \(v(t) = -32t\), substitute \(t = 3.34\) to find the velocity at impact: \(v = -32(3.34) \approx -106.88\).

Key Concepts

Velocity Function

In physics, the velocity function describes how the velocity of an object changes over time. For free-falling bodies, like Galileo's cannonball dropped from the Tower of Pisa, velocity is crucial to find out how fast the object is moving at any given time during its descent. The object is subject to constant acceleration due to gravity.
To find the velocity function of a free-falling object, we calculate the derivative of the height function concerning time. Given the height function as \(s = 179 - 16t^2\), the velocity function \(v(t)\) can be derived by taking the derivative:

• \(v(t) = \frac{ds}{dt} = -32t\).

This function tells us that the velocity increases as time passes during the fall, due to gravity. Negative sign indicates the direction is downward.

Acceleration Due to Gravity

The acceleration due to gravity, often represented by the symbol \(g\), is the constant rate at which the velocity increases for a freely falling object. On Earth, this value is approximately \(9.8\text{ m/s}^2\), but in this exercise, presented through the formula \(s = 179 - 16t^2\), it implies an acceleration scaled to \(-32\text{ ft/s}^2\).
Acceleration, in a free fall, is the derivative of the velocity function with respect to time. Using our previous velocity function \(v(t) = -32t\), we can derive the acceleration:

• \(a(t) = \frac{dv}{dt} = -32\).

This means the cannonball experiences a constant acceleration pulling it downward, unaffected by other forces like air resistance under ideal conditions.

Derivative of a Function

Derivatives play a key role in determining how a quantity changes. In the context of free fall, the derivative of a function helps us get from the position function to the velocity function, and from the velocity function to the acceleration.
To find a derivative, we take the limit of the average rate of change of the function over an infinitesimally small interval. For our example height function \(s(t) = 179 - 16t^2\), the derivative concerning time \(t\) yields:

• The first derivative is \(v(t) = -32t\), which tells us about the velocity.
• The second derivative \(a(t) = -32\), which gives us acceleration.

Understanding derivatives is essential for analyzing motion in physics and calculating instant rates of change, such as those experienced in free fall.

Impact Velocity

Impact velocity refers to the velocity of an object at the point it strikes another surface. For a free-falling object, like our hypothetical cannonball, the impact velocity is a critical factor as it indicates how fast the object is moving just before it hits the ground.
We calculate this by substituting the time at which the object reaches the ground into the velocity equation. The time \(t\) when the ball hits the ground is approximately \(3.34\) seconds, found by solving \(16t^2 = 179\). Using the velocity function \(v(t) = -32t\):

• At impact, \(v(3.34) = -32 \times 3.34 \approx -106.88\text{ ft/s}\).

The negative sign indicates the direction of motion is downward. Understanding impact velocity is crucial in predicting the force of impact and potential damage during collisions.