Posted by **shaknocka** on Monday, February 7, 2011 at 11:37am.

A dynamite blast at a quarry launches a rock straight upward, and 2.0 s later it is rising at a rate of 15 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b)4.2 s after launch.

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**shaknocka**, Thursday, February 10, 2011 at 11:46am
i found the the speed for (a) of when it was launch.. i calculate it from the formula .. v(t)= V+g(t)

V=15+9.8(2)=34.6

but how do i calculate it for (b) i tried plugging in 4.2 but i didnt work

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**shaknocka**, Wednesday, February 23, 2011 at 5:16pm
how do i find the velocity 4.2s after lunch

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**Anonymous**, Thursday, September 8, 2011 at 12:22am
It should be the absolute value of v=15-9.8*(4.2)

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