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A cannon fires a shell at an angle of 20º to the horizontal. The shell reaches a range of 436 meters, landing at the same elevation as the starting point. What was the initial velocity of the shell

  • physics -

    Range = (Vo^2/g)sin40 = 436 m

    Solve for Vo

    Yes, that is sin 40, not sin 20 in the equation. That is because of the trigonometric identity
    2 sinx cosx = sin(2x)
    Derive the range and you will see why both sinx and cosx appear.

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