Question

An object is launch straight up into the air from the ground with an initial velocity of 30 m per second object rises to highest point approximately 45 metre above the ground in 3 seconds it then falls back to the ground in three more second impacting with a speed of 30 metre per second determine the closest value of average speed and average velocity during 6s interval

Answer 1

Average Speed -

V = total distance / total time

     =  distance = gt^2 ( t = time to reach highest point = 3 seconds)

     =   10 * 9 = 90

V = 90 / 6

  = 15 m/s

Average velocity -

V = total displacement/ Total time

displacement = 0 as final and initial position are same.

v = 0