The Concept of Integral Question 5, Instructor-created question HW Score: $41.67 \%, 3.33$ of $8 \mathrm{pc}$ Points: 0 of 1 A projectile is fired in the air from a height of 125 feet. Its velocity (in feet per second) $t$ seconds after launch is $\mathrm{f}(\mathrm{t})$, given in the following table. \begin{tabular}{cccccccc} $\mathrm{x}$ & 0 & 5 & 10 & 15 & 20 & 25 & 30 \\ $\mathrm{f}(\mathrm{x})$ & 230 & 205 & 198 & 163 & 146 & 122 & 102 \end{tabular} a) How fast is the projectile traveling 15 seconds after launch? $\square$ feet per second. b) How far does the projectile move between 5 seconds and 25 seconds? Estimate $\square$ feet. c) How far does the projectile move between 0 seconds and 15 seconds? Estimate $\square$ feet. d) How high is the projectile at 25 seconds? It is $\square$ feet. (If necessary, round to two decimal places.)

Step 1 ：The velocity of the projectile 15 seconds after launch is \(163\) feet per second.

Step 2 ：To estimate the distance the projectile moves between 5 seconds and 25 seconds, we can use the average velocity over this time period and multiply it by the time interval. The average velocity is the average of the velocities at 5 seconds and 25 seconds, which is \((205 + 122) / 2 = 163.5\) feet per second. The time interval is \(25 - 5 = 20\) seconds. So the estimated distance is \(163.5 * 20 = 3270\) feet.

Step 3 ：To estimate the distance the projectile moves between 0 seconds and 15 seconds, we can use the average velocity over this time period and multiply it by the time interval. The average velocity is the average of the velocities at 0 seconds and 15 seconds, which is \((230 + 163) / 2 = 196.5\) feet per second. The time interval is \(15 - 0 = 15\) seconds. So the estimated distance is \(196.5 * 15 = 2947.5\) feet.

Step 4 ：The height of the projectile at any time is the initial height plus the integral of the velocity from 0 to that time. The initial height is \(125\) feet. The integral of the velocity from 0 to 25 seconds can be estimated as the average velocity times the time interval, which we calculated in part b) as \(3270\) feet. So the estimated height of the projectile at 25 seconds is \(125 + 3270 = 3395\) feet.

Step 5 ：Final Answer: \(\boxed{163, 3270, 2947.5, 3395}\)