A soccer player can kick the ball 29 m on level ground, with its initial velocity at 42 ∘ to the horizontal.?

At the same initial speed and angle to the horizontal, what horizontal distance can the player kick the ball on a 20 ∘ upward slope?

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  • NCS
    Lv 7
    5 months ago

    over level ground,

    range x = V²sin(2Θ)/g

    29 m = V² * sin84º / 9.8m/s²

    solves to

    V = 16.9 m/s

    Now use the trajectory equation:

    y = h + x·tanΘ - g·x² / (2v²·cos²Θ)

    where y = height at x-value of interest = x*sin20º = 0.3420x

    and h = initial height = 0 m

    and x = range of interest = ???

    and Θ = launch angle = 42º

    and v = launch velocity = 16.9 m/s

    Substituting values (and dropping units for ease -- x is in meters

    0.342x = 0 + x*tan42º - 9.8x² / (2*16.9²*cos²42º)

    0 = 0.558384x - 0.031065x²

    0 = x*(0.558384 - 0.031065x)

    which has a non-trivial solution at

    x = 18 m

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  • 5 months ago

    Question has been asked/answered before (but with different values). Take a look to see the method.

    https://answers.yahoo.com/question/index?qid=20150...

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