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An arrow is shot at 29.0° above the horizontal. Its initial speed is 54 m/s and it hits the target.

(a) What is the maximum height the arrow will attain?

(b) The target is at the height from which the arrow was shot. How far away is it?

  • Physics - ,

    at the max height, velocity vertical is zero.
    Vf=0=54*sin29deg - g*t
    solve for t. then put it in this:
    hmax=54*sin29*t-1/2 g t^2

    distance target:= 54cos29*(2t) where t is the time to the max height, 2t is the total time in the air.

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