Physics repost
posted by Lindsay on .
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?

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

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

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...