A hot-air balloonist, rising vertically with a constant velocity of magnitude v

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 4.10 m/s, releases a sandbag at an instant when the balloon is y = 40.0 m above the ground. After it is released, the sandbag is in free fall. Compute the position and velocity of the sandbag 0.250 s after its release Compute the position and velocity 1.00 s after its release How many seconds after its release will the bag strike the ground? With what magnitude of velocity does it strike? What is the greatest height above the ground that the sandbag reaches?

Explanation / Answer

Let the bag take t sec to gain its maximum height (h)
=>By v = u - gt
=>0 = 4.10 - 9.8 x t
=>t = 0.418 sec
& by v^2 = u^2 - 2gh
=>0 = (4.10)^2 - 2 x 9.8 x h
=>h = 0.858m
(a) By v = u - gt
=>v =4.10 - 9.8 x 0.25
=>v = 1.65 m/s (upward)
By s = ut - 1/2gt^2
=>h = 4.10 x 0.25 - 1/2 x 9.8 x (0.25)^2
=>h = 0.71 m
Thus position = 40 +0.71 = 40.71m (from ground)

(a)(ii) Let it gain velocity v after falling from (40 + 0.858 = 40.858 m in (1-0.418 =) 0.582 sec
Thus by v = u + gt
=>v = 0 + 9.8 x 0.582 = 5.70 m/s
& by s = ut + 1/2gt^2
=>h = 0 + 1/2 x 9.8 x (0.582)^2
=>h = 1.65 m
=>Thus position = 40.858 - 1.65 = 39.208 m(from ground)

(b) Let it take t sec to fall 40.85 8m
=>By s = ut + 1/2gt^2
=>40.858 = 0 + 1/2 x 9.8 x t^2
=>t = 2.88 sec
Time to strike ground after release (T) = 0.418 + 2.88 = 3.298 sec

(c) By v = u + gt
=>v = 0 + 9.8 x 2.88
=>v = 28.22 m/s

(d) 40.858 m

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