physics
posted by shaknocka .
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.

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 didn't work 
how do i find the velocity 4.2s after lunch

It should be the absolute value of v=159.8*(4.2)
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