1. The problem statement, all variables and given/known data A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.1 s later it is rising at a speed of 19 m/s. Assuming air resistance has no effect on the rock, calculate its speed at (a) at launch and (b) 5.2 s after the launch. 2. Relevant equations v = v0 + at (I'm assuming is the only relevant one, although I'll post two others in case they are needed) x = x0 + v0t + a/2 t^2 v^2 = v0*2 + 2a(x-x0) 3. The attempt at a solution I solved part A by plugging into the equation v = v0 + at 19 m/s = v0 + (-9.8 m/s^2)(2.1 s) And I found that the initial velocity equals 40 m/s. So, to solve part b, I should just have to plug the initial in and find the final. I tried that: v = 40 m/s + (-9.8 m/s^2)(5.2 s) v = -11 m/s However, when I entered that solution in for the homework, I was told it was wrong. I'm not really sure how to go about doing the problem if that's incorrect. I thought maybe I could have an error in rounding with significant figures. Thanks for the help!