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A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?

  • Physics -

    (Average vertical velocity) * (time of flight to max height) = Maximum height H

    (Vo*sinA/2)*(VosinA/g) = H
    Vo^2*sin^2A/(2g) = H
    Solve for Vo

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