Kylie asked in Science & MathematicsPhysics · 9 months ago

Physics Homework Help?

A ball is thrown vertically upwards in the air with an initial speed of 10 m/s and from a point that is 4.8 m above the ground. How long does it take for the ball to hit the ground? Take g = 10 m/s2 and ignore air resistance. 

2 Answers

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  • oubaas
    Lv 7
    9 months ago

    in two steps :

    Δh = Vo^2/2g = 10^2/20 = 5 m 

    tup = Vo/g = 10/10 = 1.000 sec 

    tdown = √ 2(h+Δh)/g = √ (5+4.8)*2/10 = 1.400 sec 

    t = tup + tdown = 1.00+1.40 = 2.400 sec 

    in one step

    (0-4.8) = Vo*t-g/2*t^2

    -4.8-10t+5t^2 = 0 

    t = (10+√ 10^2+96)/10 = (10+14)/10 = 24/10 = 2.400 sec 

  • 9 months ago

    Given h = 4.8 m, u = -10 m/s & g = 10 m/s^2, let the ball hit the ground after t sec,

    By s = ut + 1/2gt^2

    =>4.8 = - 10t + 1/2 x 10 x t^2

    =>5t^2 - 10t - 4.8 = 0

    =>t = [-(-10) +/- √{(-10)^2 - 4 x 5 x (-4.8)}]/[2 x 5]

    =>t = [10 +/- 14]/10

    =>t = 2.4 sec

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