Physics

A stone is thrown vertically upward at a speed of 27.70 m/s at time t=0. A second stone is thrown upward with the same speed 2.140 seconds later. At what time are the two stones at the same height?
At what height do the two stones pass each other?
What is the downward speed of the first stone as they pass each other?

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  1. At the top point v=0 =>
    time of upward motion of the 1st stone is
    t=v/g =22.7/9.8=2.32 s.

    Δt =2.32 -2.14 =0.18 s.
    At t=2.32 s the 1st stone is at the top point
    h₁=v₀t-gt²/2 =27.7•2.32 – 9.8•2.32²/2 =37.9 m
    with v=0 and begins its free fall
    The 2nd stone was at the point
    h₂= v₀Δt-g(Δt)²/2 =27.7•0.18 – 9.8•0.18²/2 =4.83 m
    and moved with upward velocity
    v₂=v₀-gΔt=
    =27.7-9.8•0.18=25.9m/s
    Δh=37.9-4.83=33.07 m

    h₁=gt²/2
    h₂=v₂t- gt²/2
    Δh= h₁+h₂=gt²/2+ v₂t- gt²/2 =v₂t
    t= Δh/ v₂=33.07/25.9=1.28 s
    h₁=gt²/2 =9.8•1.28²/2=8.03 m
    The height two stones pass each other is
    h₀= 37.9-8.03 =29.9 m
    The time of the 1st stone motion is
    2.32+1.28 = 3.6 s.
    The downward speed of the first stone is
    v=gt=9.8•1.28 =12.54 m/s.

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