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BTech, Mechanical

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A ball is thrown vertically upward with velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. At what position above the ground will they meet?

  • BTech, Mechanical -

    For the 1st ball, the time of the upward motion
    t= v₀/g =18/9.8=1.83 s.
    The height of the 1st ball
    h= v₀t-gt²/2=18•1.83 - 9.8•(1.83)²/2= =16.53 m
    The time of the 1st ball downward motion before the 2nd ball begins to move is
    Δt=2 s -1.83 s = 0.17 s
    The 1st ball during 0.17 s covered the distance
    Δh=g(Δt)²/2 =9.8•0.17²/2 = 0.14 m.
    Its downward velocity is
    v₀₁ = gΔt=9.8•0.17 =1.67 m/s
    Now, two balls begin to move:
    the 1st ball moves downward with initial velocity v₀₁ =1.67 m/s,
    the 2nd ball moves upward with initial velocity v₀₂=13.5 m/s.
    The distance separated them is
    h₀ = h- Δh = 16.53 – 0.14 = 16.39 m.
    Before the meeting, the 1st ball covered h₁=v₀₁t+gt²/2,
    and the 2nd ball covered the distance h₂=v₀₂t-gt²/2.
    h₀ = h₁+h₂=v₀₁t+gt²/2 + v₀₂t-gt²/2=
    =(v₀₁+ v₀₂)t
    t= h₀/(v₀₁+ v₀₂)=16.39/(13.5+1.67) =
    =1.08 s.
    The position above the ground is
    h₂=v₀₂t-gt²/2 =13.5•1.08 – 9.8•1.08²/2 =
    = 14.58 -5.72 = 8.86 m

  • BTech, Mechanical -

    A 1500 kg automobile is traveling up to 20 degree incline at a speed of 6 m/s. If the driver wishes to stop his car in a distance of 5m, determine the frictional force at pavement which must be supplied by rear wheels.

  • BTech, Mechanical -

    you can just solve to see when the heights are equal:

    18t-4.9t^2 = 13.5(t-2)-4.9(t-2)^2
    t = 1.08

    proceed from there as above

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