Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

Motion Along a Straight Line ?

Jack drops a stone from rest off of the top of a bridge that is 20.6 m above the ground. After the stone falls 7.0 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

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  • 1 month ago

    When all motion is going downwards it would make sence so have this direction as positive, but if that's what you require then so be it.

    -20.6 = (1/2)(-9.8)t²....using s = ut + (1/2)at² 

    so, t² = 41.2/9.8

    i.e. t = 2.05 seconds to hit the ground

    When the stone has travelled 7 metres we have,

    -7 = (1/2)(-9.8)t²

    so, t² = 14/9.8

    i.e. t = 1.2 seconds

    Hence, time taken to fall from 7 metres down to hit the ground is 2.05 - 1.2 = 0.85

    Now, 0.85 seconds is the total time taken by Jill's rock to hit the ground

    so, -20.6 = 0.85u - 4.9(0.85)²

    i.e. -20.6 = 0.85u - 3.54025

    => u = -17.05975/0.85

    so, u = -20.1 m/s....i.e. 20.1 m/s downwards     

    :)>

  • oubaas
    Lv 7
    1 month ago

    lagging time t' = √ 2h/g = √ 14/9.806 = 1.195 sec 

    dropped falling mode 

    -20.6*2 = -9.806*t^2

    t = √ 41.2/9.806 = 2.050 sec 

    thrown falling mode

    -20.6 = -Vi*(2.050-1.195) -4.903*(2.050-1.195)^2

    -20.6 = -Vi*0.855 -4.903*0.855^2

    Vi = -(20.6-3.58)/0.855 ) = -20 m/sec 

  • 1 month ago

    Jack drops a stone from rest off of the top of a bridge that is 20.6 m above the ground. After the stone falls 7.0 m, Jill throws a second stone straight down. Both rocks hit the water at the exact same time. What was the initial velocity of Jill's rock? Assume upward is the positive direction and downward is negative. (Indicate the direction with the sign of your answer.)

    how long does first rock take to fall 7.0 m

    d = ½at² + v₀t + d₀

    d = ½gt²

    t² = 7.0/4.9

    t = 1.195 sec

    the second rock takes 1.195 s to travel 20.6 m

    20.6 = 4.9(1.195)² + v₀1.195

    v₀1.195 = 20.6 – 7.0 = 13.6

    v₀ = 11.38 m/s

    Source(s): billrussell42
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