Problem 12 - Thomas Calculus

Question

Speeding bullet \(A\) 45-caliber bullet fired straight up from the surface of the moon would reach a height of \(s=832 t-2.6 t^{2}\) feet after \(t\) sec. On Earth, in the absence of air, its height would be \(s=832 t-16 t^{2} \mathrm{ft}\) after \(t\) sec. How long will the bullet be aloft in each case? How high will the bullet go?

Step-by-Step Solution

Answer

On the Moon, the bullet is aloft for 320 seconds and reaches a height of 66560 feet. On Earth, it is aloft for 52 seconds and reaches a height of 10816 feet.

Step 1: Determine the Time Aloft on the Moon

To find how long the bullet will be aloft on the Moon, set the height function for the Moon, \(s = 832t - 2.6t^2\), equal to zero and solve for \(t\). This represents when the bullet returns to the surface: \[ 832t - 2.6t^2 = 0 \] Factor out \(t\): \[ t(832 - 2.6t) = 0 \] The solutions are \(t = 0\) and \(832 - 2.6t = 0\).Solve \(832 = 2.6t\) to find the other time when the bullet is on the ground: \[ t = \frac{832}{2.6} \approx 320 \] seconds.

Step 2: Determine the Maximum Height on the Moon

The maximum height is reached when the velocity is zero. The velocity function is the derivative of the height function: \[ v(t) = \frac{d}{dt}(832t - 2.6t^2) = 832 - 5.2t \]Set \(v(t) = 0\) to find when it reaches maximum height: \[ 832 - 5.2t = 0 \] Solve for \(t\): \[ t = \frac{832}{5.2} \approx 160 \] seconds.Substitute \(t = 160\) back into the height equation to find the maximum height: \[ s = 832(160) - 2.6(160)^2 = 66560 - 66560 = 66560 \] feet.

Step 3: Determine the Time Aloft on Earth

Now, find how long the bullet will be aloft on Earth using the height function for Earth, \(s = 832t - 16t^2\). Set this equal to zero: \[ 832t - 16t^2 = 0 \]Factor out \(t\): \[ t(832 - 16t) = 0 \]The solutions are \(t = 0\) and \(832 - 16t = 0\).Find the second value of \(t\):\[ 832 = 16t \]\[ t = \frac{832}{16} = 52 \] seconds.

Step 4: Determine the Maximum Height on Earth

Find when the bullet reaches maximum height by setting the velocity (derivative) to zero:\[ v(t) = \frac{d}{dt}(832t - 16t^2) = 832 - 32t \]Set \(v(t) = 0\):\[ 832 - 32t = 0 \]Solve for \(t\):\[ t = \frac{832}{32} = 26 \] seconds.Substitute \(t = 26\) back into the height equation to find max height:\[ s = 832(26) - 16(26)^2 = 21632 - 10816 = 10816 \] feet.

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation that takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. These equations typically have two solutions, which can be found using methods like factoring, completing the square, or the quadratic formula:

• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Factoring involves rewriting the equation as \((px + q)(rx + s) = 0\) and solving for \(x\).
• Completing the square involves making a perfect square trinomial to solve for \(x\).

In the context of projectile motion, the path of an object under the influence of gravity is often modeled by a quadratic equation. For instance, in the exercise we analyzed, the height of the bullet over time on both the Moon and Earth is described by quadratic equations (\( 832t - 2.6t^2 \) and \( 832t - 16t^2 \)), helping to determine when the bullet returns to the ground (when the height \( s = 0 \)). This application highlights the importance of quadratic equations in analyzing motion.

Derivatives

Derivatives are a fundamental concept in calculus used to understand how a quantity changes. Mathematically, the derivative of a function \( f(t) \) at any point \( t \) gives the rate of change of \( f \) with respect to \( t \). Derivatives are powerful when analyzing projectile motion because they provide insight into velocity and acceleration.For example, the derivative of the height function with respect to time gives us the velocity function. In the original exercise, we found the velocity of the bullet by taking the derivative of the height functions for the Moon and Earth:

• For the Moon: \( v(t) = \frac{d}{dt}(832t - 2.6t^2) = 832 - 5.2t \)
• For Earth: \( v(t) = \frac{d}{dt}(832t - 16t^2) = 832 - 32t \)

Setting these velocity functions to zero helps find when the projectile reaches its highest point, as the velocity is zero at the peak of the motion. Thus, derivatives not only allow us to find velocity but also assist in determining critical points like maximum height.

Maximum Height

Calculating the maximum height of a projectile is an essential aspect of studying its motion. The maximum height occurs when the projectile's upward velocity reaches zero, turning into downward motion. This inflection point can be found using the derivative of the height function.For the bullet on the Moon, we determined the time \( t \) when velocity \( v(t) = 0 \) as:

• \( t = \frac{832}{5.2} \approx 160 \) seconds

At this time, plugging \( t = 160 \) into the height function gives the maximum height:

• \( s = 832(160) - 2.6(160)^2 = 66560 \) feet

Similarly, the maximum height on Earth is calculated by finding \( t \) when velocity \( v(t) = 0 \):

• \( t = \frac{832}{32} = 26 \) seconds
• Maximum height: \( s = 832(26) - 16(26)^2 = 10816 \) feet

Understanding the concepts of how the maximum height is calculated can be greatly beneficial in solving real-world problems, as maximum height defines the peak of flight in projectile motion.

Velocity and Acceleration

Velocity and acceleration are key concepts when studying the motion of projectiles. Velocity is the rate at which a projectile's position changes over time, calculated as the derivative of the height function. Acceleration, on the other hand, is the rate of change of velocity over time, determined by the second derivative of the height function.In the projectile motion exercises:

• The velocity functions were \( v(t) = 832 - 5.2t \) on the Moon and \( v(t) = 832 - 32t \) on Earth.
• Acceleration on the Moon is \( a(t) = \frac{d^2}{dt^2}(832t - 2.6t^2) = -5.2 \), which is constant and indicates gravity's pull.
• Acceleration on Earth is \( a(t) = \frac{d^2}{dt^2}(832t - 16t^2) = -32 \), showing Earth's stronger gravity compared to the Moon.

These concepts help predict how quickly a projectile slows down, stops, and begins to descend. Grasping the constant values of acceleration and how they affect motion is crucial for predicting the trajectory and behavior of projectiles under different gravitational conditions.