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Posted by on Wednesday, May 11, 2011 at 11:58am.

If a projectile has a launching angle of 52 degrees about the horizontal and an initial speed of 18 m/s, what is the highest barrier that the projectile can clear?

  • Physics - , Thursday, May 12, 2011 at 5:27pm

    Vo = 18m/s @ 52 deg.

    Vo(hor) = 18cos52 = 11.08m/s.

    Vo(ver) = 18sin52 = 14.18m/2,up.

    Vf^2 = Vo^2 + 2gd = 0 @ max height.
    (14.18)^2 + 2(-9.8)d = 0,
    201.07 -19.6d = 0,
    d = 10.3m. = max barrier that can be
    cleared.

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