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## Will mark as BRAINLIEST-.. A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated a

Question

Will mark as BRAINLIEST……

A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated at a height of 100m.

How long does it take the packet to reach the ground ?

What is it’s final velocity ?

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Physics
6 months
2021-08-10T20:08:20+00:00
2021-08-10T20:08:20+00:00 1 Answers
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## Answers ( )

Answer:t = 5.04secsv = –44.49m/sExplanation:Given parametersinitial velocity u = 4.9m/s

height of drop H = 100m

acceleration due to gravity g = -9.8m/s²(since the body is ascending i.e moving against gravity)

Requiredtime taken by the packet to reach the ground ‘t’

Using the equation of motion

S = ut+1/2gt²0-100 = 4.9t+1/2(-9.8)t²

-100 = 4.9t-1/2(9.8)t²

-100 = 4.9t-4.9t²

4.9t-4.9t² +100 = 0

4.9t²-4.9t² -100 = 0

Multiplying through by 10

49t²-49t-1000 = 0

Using the general equation t = (-b±√b²-4ac)/2a

t = -(-49)±√(-49)²-4(49)(-1000)/2(49)

t = 49±√(2401+196000)/2(49)

t = 49±√(198401)/98

t = 49±√(198401)/98

t = 49±449.42/98

t = (49+449.42)/98

t = 494.42/98

t = 5.04secsHence, it took the packet 5.04secs to reach the groundThe final velocity is calculated using the formula v = u + at

v = u-gtv = 4.9-(9.8)(5.04)

v = 4.9-49.392

v = -44.492m/sThis shows that the final velocity is 44.49m/s(moving downwards)