Determine the velocity at which the bullet-block system hits the ground.

[tawko]

1. A bullet of mass 0.0020 kg and traveling at 300 m/s strikes the center of a 3.0 kg block of wood which is sitting on a fence post. The block of wood is 2.0 m above the ground.
I got the velocity using the eq (m1v1 + m2v2)/Mtotal = 0.2 m/s [correct if incorrect please]

I then got the horizontal distance by finding time and using x = v*t

Determine the velocity at which the bullet-block system hits the ground. Be sure to include an angle.
2. a^2 + b^2 = c^2 Vfinal = at + Vinitial Xfinal = 1/2at (m1v1 + m2v2) / Mtotal
3. I figured this would involve a right triangle, so I tried solving for the length of the sides. I believe this is where I went wrong. The height of the triangle is 6.272 because i used Vfinal=at+Vinitial, 9.8(.64)+0. I then used the Pythagorean theorem but the answer seemed strange. Thanks.

 

[Mindscrape]

You did it right as far as I can tell, though you don't need the horizontal distance for anything (don't give yourself more work than necessary). By legs of the right triangle, you mean the horizontal velocity and the vertical velocity, right? If so, you're good. :)