Question

Part A Constants Compute the position of the sandbag at 0.250 s after its release 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 Express your answer in meters to three significant figures Submit Request Answer Figure 1 of 1 5.00 ms Part B Compute the position of the sandbag 1.08 s after its release Express your answer in meters to three significant figures 40.0 m to ground Submit Request Answer

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Answer 1

U = 5 m/s S_0 = 40 m a = negative 9.8 m/s^2 A) = h = s_0 + ut + 1/2 at^2 = 40 + 5 times 0.25 - 9.8/2 times 0.25^2 = 40.94375 m = 40.9 m h = s_0 + u t + 1 /2 at^2 = 40 + 5 times 1.08 - 9.8 /2 times 1.08^2. = 39.685 m = 39.7 m \r\n c) V = u + at = 5 + (negative 9.8) (0.25) = 2.55 m/s D) v = u + at = 5 - 9.8 times 1.08 = negative 5.584 m /s
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