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.
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)
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!