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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Thu Cúc 6 months 2021-08-10T20:08:20+00:00 1 Answers 8 views 0

Answers ( )

    0
    2021-08-10T20:10:07+00:00

    Answer:

    t = 5.04secs

    v = –44.49m/s

    Explanation:

    Given parameters

    initial 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)

    Required

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

    Using the equation of motion

    S = ut+1/2gt²

    0-100 = 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 = 5.04secs

    Hence, it took the packet 5.04secs to reach the ground

    The final velocity is calculated using the formula v = u + at

    v = u-gt

    v = 4.9-(9.8)(5.04)

    v = 4.9-49.392

    v = -44.492m/s

    This shows that the final velocity is 44.49m/s(moving downwards)

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