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A stone is dropped into a river from a bridge 43.3 m above the water. Another stone is thrown vertically down 1.06 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

  • physics -

    h=gt²/2 =>
    t= sqrt(2h/g) = sqrt(2•43.3/9.8) =2.97 s.
    t1=t-1.06 = 2.97 – 1.06 = 1.91 s.
    h= v₀•t1 +g•t1²/2
    Solve for v₀.

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