A person standing on the roof of a building throws a ball directly upward. The ball misses the rooftop on its way down andeventually strikes the ground. The function5(1) - - 16:1 + 64! + 80describes the ball’s height above the ground, sit), in feet, tseconds after it was thrown. a. Fund the balls average velocity between the time it was thrown and 2 seconds later. b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown. c. What do the signs in your answers to parts (a) and (b) mean in terms ofthe direction of the ball’s motion?

a) V_avg = 32 ft/s

b) V_avg = -30 ft/s

c) See explanation

Step-by-step explanation:

Given:

- The displacement function is given:

                                 S(t) = -16t^2 + 64t + 80

Find:

a. Find the balls average velocity between the time it was thrown and 2 seconds later

b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown

c. What do the signs in your answers to parts (a) and (b) mean in terms of the direction of the ball’s motion?

Solution:

- The average velocity of an object is the displacement over time, and it is expressed as follows:

                                  V_avg = ( S_2 - S_1 ) / dt

- Evaluate S(2), S(0), and S(4):

                                  S(2) = -16*2^2 +64*2 +80

                                  S(2) = 144 ft

                                  S(0) = -16*0^2 +64*0 +80

                                  S(0) = 80 ft

                                  S(4) = -16*4^2 +64*4 +80

                                  S(4) = 80 ft

- For part a. Compute V_avg:

                                  V_avg = ( S(2) - S(0) ) / 2

                                  V_avg = (144 - 80) / 2

                                  V_avg = 32 ft/s

- For part b. Compute V_avg:

                                  V_avg = ( S(4) - S(2) ) / 2

                                  V_avg = (80 - 140) / 2

                                  V_avg = -30 ft/s

- We see that the signs of the Velocity have changed. The coordinate axis up against gravity was chosen to positive. While a negative access is in favor of the gravity. Hence, from the signs we can see that the ball is going down towards ground in motion from t = 2 s to t = 4 s.

stops on an even number.