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Arelis
Arelis asked in Science & MathematicsPhysics · 1 day ago

A stone that starts at rest is in free fall for 4.0 seconds. What's the stone’s velocity after 4.0 seconds?

 and stone’s displacement during this time? 

Physics

3 Answers

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  • oldschool
    Lv 7
    1 day ago

    Vf = Vi + a*t = 0 + 9.8*4 = 39m/s

    d = do + Vo*t +½*a*t² = 0 + 0*t +½*9.8*4² = 78.4 or 78m with 2 s.f. 

  • Morningfox
    Lv 7
    1 day ago

    Each second, the stone gains speed of 9.8 m/s, downwards.

  • oubaas
    Lv 7
    1 day ago

    final velocity V = g*t = -9.806*4 = -39.2 m/sec 

    displacement d = V*t/2 = g*4*4/2 = g*8 = -9.806*8 = -78.4 m/sec 

    displacement d = g/2*t^2 = g*16/2 = g*8 = -9,806*8 = -78.4 m 

    displacement d = V^2/2g = g^2*16/2g = g*8 = -9.806*8 = -78.4 m 

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