A dynamite blast at a quarry launches a rock straight upward, and 2.2 s later it is rising at a rate of 17 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b)4.4 s after launch.

Answer 1

a. V = Vo + g*t = 17 m/s.
Vo = 17-g*t = 17-(-9.8*2.2) = 38.56 m/s.

b. V = 38.56 -9.8*4.4 = -4.56 m/s.

The negative sign means the rock is on its' way down.