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Question

As a runaway scientific balloon ascends at 19.6 m/s, one of its instrument packages breaks free of a harness and free-falls. Figure below gives the vertical velocity of the package versus time, from before it breaks free to when it reaches the ground. (a) What maximum height above the break-free point does it rise? (b) How high is the break-free point above the ground?

Take g = 9.8 m/s2


A

above break free point 19.6 m

height of break free point 78.4

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B

above break free point 78.4 m,

height of break free point 19.6

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C

above break free point 19.6 m,

height of break free point 58.8

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D

above break free point 58.8 m,

height of break free point 9.8

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Solution

The correct option is C

above break free point 19.6 m,

height of break free point 58.8


Till t = 2 sec the velocity of package is same as that of balloons i.e., 19.6 m/s

As soon as the parcel leaves the balloon its velocity is + 19.6 m/s (upward). Its acceleration is - 9.8 m/s2(downwards)

Now the parcel will keep going up till its velocity becomes 0.

So v = 0 u = +19.6 a = - 9.8 m/s2

v2u2=2as

(19.6)2= 2(-9.8) s

9.8×42=s=19.6m

So, parcel goes 19.6 m above the break free point.

The parcel broke free at t = 20 sec. with v= + 19.6

The parcel hits ground at t = 8 sec

So time of flight = 8 - 2 = 6 sec

u = 19.6 t = 6 sec a = - 9.8 m/s2

s=19.6×6+12+(9.8)(6)2

s=58.8m

Negative sign says that ball was displaced by 58.8m downwards from the breakaway point.


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