Physics repost

A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground?

I keep getting 64.28 m/s as my answer, but it's not right. Could someone tell me exactly what I’m doing wrong?

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  1. Final KE=1/2 m vi^2 + mgh
    = 1/2 m 41^2 + m*9.8*190
    1/2 m vf^2= 1/2 m( 3.55E5)

    vf= 59.5m/s

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    bobpursley
  2. Would 59.5 be the final answer?? B/c that is not right, either...

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  3. Check my work. The initial velocity is given as two significant digits, so the final should be in ... digits.

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    bobpursley
  4. I keep getting the same answer as you.

    This is what I did to get my original answer:
    V^2 = sqrt[Vx^2 + Vy^2] where...

    Vx = 41 m/s * cos 38
    Vy at impact can be calculated using
    Vy^2 = (41 sin 38)^2 + 2 g H
    I got 55.56 as my Vy. I have a feeling this is where I got messed up...

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