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Posted by on Sunday, September 5, 2010 at 12:35pm.

at what projection will the range of a projectile equal to 2 times its maximum height?

  • phy - , Sunday, September 5, 2010 at 12:50pm

    The time of flight is given by 2Vi sinTheta/g

    Range is then Vi*t*costheta=2Vi^2sinThetaCosTheta/g

    The maximum height is given by H = (v sin θ)^2 / (2 g).

    work all those out yourself to verify true.
    Now, Range=2Height

    2Vi^2sinThetaCosTheta/g =2(Vi*sinTheta)^2/2g
    reducing that,

    sinTheta/cosTheta=2
    theta=arctan 2

    check my math.

  • phy - , Sunday, September 5, 2010 at 1:11pm

    thanks

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