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posted by Lindsay on Thursday, November 8, 2007 at 9:51pm.

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? Which equations do I need for this?

V^2 = sqrt[Vx^2 + Vy^2] Vx remains 41 m/s * cos 38 during flight. Vy at impact can be calculated using Vy^2 = (V sin 38)^2 + 2 g H

V is 41 m/s, right?

I keep getting 64.28 m/s as my answer, but it's not right. What am I doing wrong?

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