Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 61: 16

a). Net displacement of the ball = $90.55 ft$ b). Average velocity = $=36.22 ft/s$ c). We cannot calculate the average speed, since we do not know the trajectory of the ball in air.

1 yd = 3 ft So, 30 yd = 90 ft And height of goal = 10 ft a). Net displacement of the ball = $\sqrt (90^{2}+10^{2})=\sqrt (8100+100)=\sqrt 8200=90.55 ft$ b). Average velocity = $\frac{90.55ft}{2.5s}=36.22 ft/s$ c). We cannot calculate the average speed, since we do not know the trajectory of the ball in air.