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A ball thrown into the air lands on the same horizontal level, 31 m away, and 2.65 s later. Find the magnitude of the initial velocity.

  • Physics - ,

    if it rose and fell for 1.325s, then it rose and fell 4.9*1.325^2 = 8.6m.

    Now, for a given angle θ and initial velocity v,

    range R = v^2 sin(2θ)/g
    max height H = v^2 sin^2(θ)/g

    R/H = 4cotθ
    So, here, 31/8.6 = 4cotθ
    cotθ = 0.90
    θ = 48°

    using the range R,

    31 = v^2 sin96°/g
    v^2 = 31g/sin96° = 305.5
    v = 17.48

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