Physics - check my work

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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?

Please check my work. I think it’s wrong, but I don’t know where I’m messing up…

Vx = 41 m/s * cos 38 = 32.3 m/s
Vy at impact can be calculated using Vy^2 = (41 sin 38)^2 + 2 g H
Vy = 55.56 m/s (I think this is the part that’s wrong)

V^2 = sqrt[Vx^2 + Vy^2]
V^2 = 64.27 m/s

  • Physics - check my work -

    The formula you used to get vyf is from energy. Energy is not a vector, so you cannot use that equation to get vyf, a vector.

  • Physics - check my work -

    Hmm yeah I figured Vfy was wrong. But I'm still a little confused on which I equation I DO need to use.

  • Physics - check my work -

    I finally got the right answer. And I was using the right equation for Vfy, just not using the right numbers.

  • Physics - check my work -

    Actually the energy equation CAN be applied using Vy only, because the Vx component is constant. It is a handy shortcut. You can think of it as an energy equation being applied in an intertial coordinate system moving with the lateral velocity of the ball.

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