Posted by **Lindsay** on Friday, November 9, 2007 at 8:01am.

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**, Friday, November 9, 2007 at 8:24am
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**, Friday, November 9, 2007 at 8:32am
Would 59.5 be the final answer?? B/c that is not right, either...

- Physics repost -
**bobpursley**, Friday, November 9, 2007 at 8:44am
Check my work. The initial velocity is given as two significant digits, so the final should be in ... digits.

- Physics repost -
**Lindsay**, Friday, November 9, 2007 at 9:02am
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...

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