Problem 30 - Thomas Calculus

Question

Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater's floor, which at one point shot lava 1900 \(\mathrm{ft}\) straight into the air (a world record). What was the lava's exit velocity in feet per second? In miles per hour? (Hint: If \(v_{0}\) is the exit velocity of a particle of lava, its height \(t\) sec later will be \(s=v_{0} t-16 t^{2} \mathrm{ft} .\) Begin by finding the time at which \(d s / d t=0 .\) Neglect air resistance.)

Step-by-Step Solution

Answer

The lava's exit velocity is 348 ft/s or 237.27 mph.

Step 1: Understand the Problem

We need to find the exit velocity of the lava. The hint given suggests using the height formula: \[s = v_0 t - 16t^2\]This formula models the height of the lava stream over time. Our task is to find the maximum height and use it to determine the exit velocity \(v_0\).

Step 2: Differentiate to Find Maximum Height

Differentiate the height function \(s = v_0 t - 16t^2\) with respect to \(t\) to find when the height is maximized: \[\frac{ds}{dt} = v_0 - 32t\]Set \(\frac{ds}{dt} = 0\) to find the time at which the maximum height occurs: \[v_0 - 32t = 0\]

Step 3: Solve for Time at Maximum Height

From the equation \(v_0 - 32t = 0\), solve for \(t\): \[v_0 = 32t\]\[t = \frac{v_0}{32}\]

Step 4: Use Maximum Height to Determine Exit Velocity

At maximum height, the lava's height is \(1900\) feet. Substitute \(t = \frac{v_0}{32}\) back into the height equation:\[1900 = v_0 \left(\frac{v_0}{32}\right) - 16\left(\frac{v_0}{32}\right)^2\]Solve this equation to find \(v_0\).

Step 5: Calculate Exit Velocity in Feet per Second

Simplify the expression:\[1900 = \frac{v_0^2}{32} - \frac{16v_0^2}{1024}\]Combine terms:\[1900 = \frac{v_0^2}{32} - \frac{v_0^2}{64}\]\[1900 = \frac{2v_0^2 - v_0^2}{64}\]\[1900 = \frac{v_0^2}{64}\]Solve for \(v_0\):\[v_0^2 = 121600\]\[v_0 = \sqrt{121600} = 348\text{ ft/s}\]

Step 6: Convert Exit Velocity to Miles Per Hour

Convert \(348\) feet per second to miles per hour using the conversion factor: \[1 \text{ mile} = 5280 \text{ feet},\]\[1 \text{ hour} = 3600 \text{ seconds}.\]Calculate:\[v_0 = 348 \times \frac{3600}{5280} = 237.27 \text{ mph}\]

Key Concepts

Differentiation

Differentiation is a cornerstone of calculus that allows us to determine how a function changes at any point. In the context of this exercise, we are interested in the rate of change of the lava's height over time. This is described by the derivative of the height function.
The height of the lava is given by the function:
\[ s = v_0 t - 16t^2 \]
Here, differentiation helps us find the first derivative of the height function with respect to time, which is:
\[ \frac{ds}{dt} = v_0 - 32t \]
This derivative represents the velocity of the lava at any moment in time. By setting this derivative equal to zero, we find the point in time at which the height is maximized. This is because the rate of change in height is zero at the maximum height. Understanding how differentiation relates to the change in height is crucial to solving this problem effectively.

Maximum Height

The maximum height of the lava occurs when its upward velocity is zero. This means that for a brief moment, the lava stops rising before descending again.
After differentiating, we set the first derivative equal to zero to find this critical point:
\[ v_0 - 32t = 0 \]
Solving for \(t\), we find:
\[ t = \frac{v_0}{32} \]
This gives us the time at which the lava reaches its peak height. At this time, the given maximum height of the lava is 1900 feet.
Substituting back into the equation allows us to relate maximum height directly to the exit velocity.

Velocity Calculation

The exit velocity is the initial velocity at which the lava leaves the volcano. From the formula:
\[ 1900 = \frac{v_0^2}{32} - \frac{v_0^2}{64} \]
we can solve for the exit velocity \(v_0\).
Simplifying the equation step-by-step leads to:

• Combine like terms to simplify the equation.
• Solve for \(v_0^2\), which results in 121600.
• Taking the square root gives us the exit velocity in feet per second \(v_0 = 348\ \text{ft/s}\).

This conversion provides a clear numerical depiction of how fast the lava bursts out initially.

Kinematic Equations

Kinematic equations describe the motion of objects and are particularly useful in physics.
In this exercise, the height equation \( s = v_0 t - 16t^2 \) is a kinematic equation modeling projectile motion under gravity.

• \(v_0 t\) represents the initial motion due to the exit velocity.
• The term \(-16t^2\) accounts for the gravitational pull acting downward on the lava.

Understanding how the interplay of initial velocity and gravitational force affects an object is key to calculating motion parameters such as velocity and maximum height. These equations allow us to deduce real world phenomena from mathematical models efficiently.