Question

Vertical Motion In Exercises 97 and \(98,\) use the position function \(s(t)=-16 t^{2}+v_{0} t+s_{0}\) for free-falling objects. A ball is thrown straight down from the top of a 220 -foot building with an initial velocity of \(-22\) feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?

Short answer

The velocity of the ball after 3 seconds is -118 feet/second, and the velocity of the ball after falling 108 feet is approximately -77.68 feet/second.

Step-by-step solution

Deriving the velocity function

The position function for this problem is given as \(s(t)=-16t^{2}+v_{0}t+s_{0}\) with \(v_0 = -22\) feet/second (the initial velocity) and \(s_0 = 220\) feet (the initial height or position). The velocity function is the derivative of the position function: \(v(t) = s'(t) = -32t + v_0\). Substituting \(v_0 = -22\), the velocity function becomes \(v(t) = -32t - 22\).

Calculating the velocity after 3 seconds

With the derived velocity function, substitute \(t = 3\) seconds. Thus, \(v(3) = -32(3) - 22 = -96 - 22 = -118\) feet/second. So, the velocity of the ball after 3 seconds is -118 feet/second.

Finding the time when the ball has fallen 108 feet

To find the velocity after falling 108 feet, one first needs to find out how long it takes for the ball to fall this distance. This requires setting the position function equal to \(220 - 108 = 112\) feet (the remaining distance to the ground after the ball has fallen 108 feet), and solving for \(t\). Therefore, \(112 = -16t^2 - 22t + 220\). Solving this quadratic equation yields \(t \approx 1.74\) seconds.

Calculating the velocity after falling 108 feet

With the derived time, substitute \(t \approx 1.74\) seconds into the velocity function. Thus, \(v(1.74) \approx -32(1.74) - 22 = -55.68 - 22 \approx -77.68\) feet/second. Therefore, the velocity of the ball after falling 108 feet is approximately -77.68 feet/second.