Physics  check my work
posted by Lindsay .
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

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.

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

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

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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