12. A hot-air balloon is moving vertically upward with a speed of 10.0 m/s. The

Question

12. A hot-air balloon is moving vertically upward with a speed of 10.0 m/s. The pilot of the hot-air balloon gingerly unties a sandbag on the balloon at the instant that the balloon is 50.0 m above the ground, after which the sandbag is freely falling. (a) Measured relative to the ground, to what maximum height does the sandbag reach? (b) Compute the position of the sandbag at 0.500 s and 2.00 s after being released (c) For how uch time is the sandbag in flight after being released? (d) How fast is the sandbag moving immediately before hitting the ground? (e) Upon choosing an appropriate coordinate system, accurately sketch graphs of the positi acceleration of the sandbag as functions of the point immediately before it strikes the ground. the time, t, starting from when the sandbag is released up to

Explanation / Answer

Accoridng to the given problem,

a) As it's mentioned in the problem,

The maximu height of sand bag is 50m relative to ground.

b) Using the kinamatic equation of freely falling body,

S = 1/2gt2 where g = accelration due to gravity, t time of fall.

at t= 0.5s,

S = 1/2*9.81*0.52 = 3.67875m From intial drop point.

and 46.32 m relative to ground.

at t=2s,

S = 1/2*9.81*22 = 19.62 m From drop point.

and 30.38 m relative to ground.

c)Total time of free fall,

Using the same formula, but here S = 50m total height it travelled .

50 = 1/2*9.81*t2

t = 3.19275s

t = 3.2 s

d) Using the kinamtic equation of motion,

v = u + gt Here u = 0 as its a free fall and t [calculate in part c]

v = gt

v = 31.32 m/s

As per chegg guidlines only 4 sub-parts per Question.

I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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