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?

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

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

Physics repost 
Lindsay,
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...