Question

A dynamite blast at a quarry launches a rock straight upward and 1.5s later it is rising at a rate of 11m/s. Assuming air resistance has no effect on the rock calculate its speed (a) at launch and (b) 5.1s after launch. I have gotten the answer for (a) which is 25.7m/s however I can not get (b). I have tried using the V=vo+at formula to get (b) which gives you -24.28 but this answer keeps on coming up as incorrect. I’m unsure what to do next.

Answer 1

a)

$$v_i-v_f=gt$$

$$v_i-11=(9.8)(1.5)$$

$$v_i=25.7\frac{m}{s}$$

b) The velocity:

$$V=v_f-gT=11-9.8(5.1-1.5)=-24.3\frac{m}{s}$$

The speed:

$$v=24.3\frac{m}{s}$$