Question

You throw a ball upward with an initial speed of 4.8 m/s . When it returns to your hand 0.98 s later, it has the same speed in the downward direction.What was the average acceleration vector of the ball?

Answer 1

a = (V2 - V1) / t average acceleration between points 1 & 2

V2 = speed at bottom = -4.8 m/s

V1 = 4.8 m/s speed when thrown, choosing upwards as positive

a = (-4.8 - 4.8) / .98 time upwards = time downwards

a = -9.6 m/s / .98 s

a = -9.8 m/s^2