Physics

A gray kangaroo can bound across level ground with each jump carrying it 8.4m from the takeoff point. Typically the kangaroo leaves the ground at a 23 angle.
1. Provide knowns and unknown values.
2. What is the initial velocity or takeoff speed?
3. What is its maximum height above the ground?

Please provide formula and numerical answer.

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  1. say speed is S
    then horizontal velocity u = S cos 23
    Vi, initial vertical velocity = S sin 23
    t is total time in air and t/2 is time to max height

    d = distance = u t
    so
    8.4 = u t = .920 S t
    S t = 9.13

    at top
    time = t/2 = 9.13/2S = 4.57/S
    v = 0 = Vi - 9.8 (4.57/S)
    0 = .391 S - 44.8/S
    .391 S^2 = 44.8
    S = 10.7 m/s takeoff speed

    h = Vi (t/2) - 4.9 (t/2)^2
    t/2 = 4.57/S = .427
    Vi = .391 (10.7) = 4.18
    h = 4.18(.427) - 4.9 (.427)^2
    h = .891

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