Problem 43 - College Algebra and 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 function of the ball is \(y(t) = 144*t - 0.5*32.2*t^2\) feet. The velocity function is \(v(t) = 144 - 32.2*t\) feet per second, and the acceleration function is \(a(t) = -32.2\) feet per second squared. The ball reaches its highest point at \(t = 144 / 32.2 \approx 4.472s\), it reaches approximately \(y(4.472) = 144*4.472 - 0.5*32.2*4.472^2\) height. The velocity when it hits the ground is equal to the initial velocity but in the opposite direction.

Step 1: Position Function

First, we'll define the function for the position of the ball. We can calculate the position function y(t) of the ball at any time t using the initial velocity and using the equation of motion: \(y(t) = y_0 + v_0*t - 0.5gt^2\). Given, \(v_0 = 144 ft/s\), \(y_0 = 0 ft\) (since the ball is propelled from the ground) and gravitational constant \(g = 32.2 ft/s^2\) (approximately), the position function becomes \(y(t) = 144*t - 0.5*32.2*t^2\).

Step 2: Velocity and Acceleration Functions

To find the velocity function \(v(t)\) and acceleration function \(a(t)\), we get them by differentiating the position function as per calculus rules. The velocity is the first derivative of position, and acceleration is the first derivative of the velocity function or the second derivative of the position function. As such, \(v(t) = y'(t) = 144 - 32.2*t\), and \(a(t) = v'(t) = -32.2\). The acceleration becomes a constant as in case of free fall, it does not change and is always equal to the gravitational constant -32.2ft/s^2

Step 3: Finding Highest Point's Time

The highest point is reached when the velocity equals zero (\(v(t)= 0\)). Equating the velocity function to 0, \(144 - 32.2*t = 0\). Now solve this for t to get the time at which ball reaches highest point. \(t = 144 / 32.2 \approx 4.472s\).

Step 4: Finding Highest Point's Position

Plug \(t = 4.472s\) into the position function \(y(t)\) to get the highest point position: \(y(4.472) = 144*4.472 - 0.5*32.2*4.472^2\). Calculate to get the highest position.

Step 5: Finding Velocity When Ball Hits Ground & Relation to Initial Velocity

The ball hits the ground when position function \(y(t) = 0\). Solve this equation for t to get the time when the ball hits the ground. Put this value in velocity function to get the velocity at this time point. The negative sign indicates the direction is downwards. It should be equal to the initial velocity just in the opposite direction. It shows that the speed of the ball when it hits the ground is equal to the speed it was thrown up at.