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a projectile with a muzzle velocity of 550 m/s is fired from the a gun on top of a cliff 460 m above the sea level at a certain angle with respect to the horizontal. if the projectile hits the ocean surface. 49.2 seconds after being fired, determine the horizontal range of the projectile?

  • physics -

    Range = Vo^2*2sinA*cosA/g = Vo*cosA*T
    (550)^2*2sinA*cosA/9.8=550cosA*49.2
    Divide both sides by 550cosA:
    550*2sinA/9.8 = 49.2
    56.12*2sinA = 49.2
    112.24*sin A = 49.2
    sinA = 0.43833
    A = 26o.

    Range = Vo*cos A * T
    Range = 550*cos26 * 49.2 = 24321 m.

  • physics -

    Hi

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