Problem 43 - Calculus an Applied Approach

Question

A ball is propelled straight upward from ground level with an initial velocity of 144 feet per second. (a) Write the position, velocity, and acceleration functions of the ball. (b) When is the ball at its highest point? How high is this point? (c) How fast is the ball traveling when it hits the ground? How is this speed related to the initial velocity?

Step-by-Step Solution

Answer

The position, velocity, and acceleration functions are \(s(t) = 144t - 16t^2 feet\), \(v(t) = 144 - 32t ft/s\) and \(a(t) = -32 ft/s^2\). The ball reaches its highest point of 324 feet after 4.5 seconds. When the ball hits the ground after 9 seconds, it is traveling at 144 feet per second in the opposite direction to the initial velocity.

Step 1: Write the position, velocity, and acceleration functions of the ball

Given initial velocity (\(v_0\) ) as 144 feet per second and gravity as \(32ft/s^2\), the acceleration \(a(t)\) due to gravity is constant, \(a(t) = -32 ft/s^2\). Velocity \(v(t)\) is the integral of acceleration, \(v(t) = v_0 + a(t) * t = 144 - 32t ft/s\). And position \(s(t)\) is the integral of velocity, \(s(t) = s_0 + v_0*(t) + 0.5 * a(t) * t^2 = 144t - 16t^2 feet\).

Step 2: Find when the ball is at its highest point and its height

The ball is at its highest point when the velocity is zero. By setting the velocity equation \(144 - 32t = 0\), we find when velocity is zero, \(t = 4.5\), this is when ball is at its highest point. We then plug this time into the position equation to find the height, \(s(4.5) = 144(4.5) - 16(4.5)^2 = 324 feet\).

Step 3: Find how fast the ball is traveling when it hits the ground

The ball hits the ground when the position \(s(t) = 0\). Setting the position equation \(144t - 16t^2 = 0\), we find the time of when it hits the grounds is at 0 second (starts from the ground) and \(t = 9\) seconds (when it returns to the ground). We have \(v(9) = 144 - 32*9 = -144 ft/s\). In magnitude, this is the same as the initial velocity.

Key Concepts

Position Function

In projectile motion, the position function describes the height of an object over time. For a ball thrown upwards, the position function is influenced by initial velocity and gravity. Given a ball with an initial velocity of 144 feet/second and starting from ground level, the position function can be determined using:

• Initial position: 0 feet (starting from ground level)
• Initial velocity (\(v_0\)): 144 feet/second
• Acceleration due to gravity (\(a\)): -32 feet/secondy²

The position function is given by:\[s(t) = v_0 \, t - \frac{1}{2} \, a \, t^2 = 144t - 16t^2\]This formula shows how height changes with time until the ball returns to the ground.

Velocity Function

Velocity is the rate of change of position over time. For projectile motion, velocity starts at the initial speed and decreases due to gravity. The velocity function for our ball situation can be calculated as:

• Initial velocity (\(v_0\)): 144 feet/second
• Acceleration due to gravity (\(a\)): -32 feet/secondy²

The velocity function is given by:\[v(t) = v_0 - a \, t = 144 - 32t\]This shows how velocity decreases as the ball goes up, reaching zero at the peak height.

Acceleration Function

Acceleration in this context is constant and results from gravity. It tells us how quickly the velocity of an object changes. For a projectile like a ball thrown upwards:

• Acceleration due to gravity is constant at -32 feet/secondy²

The acceleration function is represented as:\[a(t) = -32\]This value remains constant, acting downward and slowing the ball's upward motion until it starts descending.

Initial Velocity

Initial velocity is the speed at which an object starts its motion. For our ball, the initial velocity is vital in determining how high and how long it will travel. In this case:

• Initial velocity (\(v_0\)): 144 feet/second

This speed sends the ball upward against gravity. Initial velocity impacts both the maximum height and total airtime of the projectile.

Maximum Height

The maximum height is the peak position the ball reaches during its flight. Achieved when the velocity becomes zero due to gravity, we use the time derived from the velocity function to determine this:

• Find time at maximum height by solving \(144 - 32t = 0\), which gives \(t = 4.5\) seconds.
• Substitute \(t = 4.5\) into the position function to find height.

The maximum height \(s(4.5)\) is calculated as:\[s(4.5) = 144(4.5) - 16(4.5)^2 = 324 \, \text{feet}\]This is the highest point the ball reaches on its trajectory.