Could someone help with this physics question? Please include work?

A projectile is to be shot at 50m/s over the level ground in such a way that it will land 200m from the shooting point. At what angle, in degrees, should the projectile be shot? 

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  • oubaas
    Lv 7
    1 month ago

    200 = 50^2/g*k 

    k = 200*9.807/2500 = 0.785 = sin 2Θ

    Θ = (arcsin 0.785) / 2 = 25.86°

  • Ash
    Lv 7
    1 month ago

    Range, R = v²sin2θ/g

    sin2θ = Rg/v²

    2θ = arcsin (Rg/v²)

    θ = ½ arcsin (Rg/v²)

    θ = ½ arcsin (200*9.8/50²)

    θ = 25.8°

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