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The highest barrier that a projectile can clear is 12.7 m, when the projectile is launched at an angle of 17.5° above the horizontal. What is the projectile's launch speed?

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The vertical initial speed is Vsin17.5

the vertical speed at max height is zero.

Vf^2=(Vsin17.5)^2-2*9.8*12.7
solve for V, knowing Vf is zero.

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