(adsbygoogle = window.adsbygoogle || []).push({}); 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!

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# Final velocity given acceleration and initial velocity

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