2.2. A hot air balloonist, rising vertically with a constant velocity of magnitude 5.00m/s releases a sandbag at an instant when the balloon is 40.0m above the ground as shown. After it is released, the sandbag is in free fall.

Question

1. calculate the position and velocity of the sandbag at 0.25s and 1.00s after its release.

Answer 1

a. h = ho - (Vo*t + 0.5g*t^2)
h = 40 - (-5*0.25 + 4.9*0.25^3) = 40.9 m
V = Vo + g*t = -5 + 9.8*0.25=-2.55 m/s.

b. h = 40 - (-5*1 + 4.9*1^2) = 40.1 m.
V = -5 + 9.8*1 = 4.8 m/s.