Question

A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s , releases a sandbag at the instant the balloon is 40.0 m above the ground. (See (Figure 1)) After it is released, the sandbag encounters no appreciable air drag.

**only need help with parts E-J

Part A Compute the position of the sandbag at 0.270 s after its release. Express your answer in meters to three significant figures. h-41.0 m Previous Answers Correct PartB Compute the position of the sandbag 1.02 s after its release. Express your answer in meters to three significant figures. h-40.0 m vious Ans Correct

Answer 1

ým
V =/ V = u + at, t_1 = 1/2 = 0.5 sec = 5 times 1/2 - 1/2 times 10 times 1/4 = 2.5 - 1.25 = 1.25 m H = 40 + 1.25 H = 41.25 m t = V = squareroot 2gr' V = squareroot 2 times 10 times 41.25 V = 28.72 m/sec t = 28.72 = 2.87 m/sec E 2.87 sec F V = 28.72 m/sec G 41.25 m \r\n