Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

Hotairballon is dropping at 2.2m/s when person drops camera. If camera drops at 40m above ground how long does it take camera to reach floor?

Also, what is its velocity just before it lands? (m/s) Let upward be positive direction

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  • Whome
    Lv 7
    1 month ago
    Favorite Answer

    s = s₀ + v₀t + ½at²

    0 = 40 +(- 2.2)t + ½(-9.8)t²

    0 = 40 - 2.2t - 4.9t²

    t = (2.2 ±√(2.2² - 4(-4.9)(40))) / (2(-4.9))

    t = -3.09 s  Ignore 

    or

    t = 2.6414...

    t = 2.6 s ANSWER

    v = v₀ + at

    v = -2.2 - 9.8(2.6)

    v = - 28 m/s

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  • 3 weeks ago

    The camera reaches ground as a horizontal projectile, as such it’s time of flight should be 

         t = square root of ( 2h/g)= sq.root( 2x40/9.8)

            = 20/7 sec = 3 sec (nearly)

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