A dynamite blast at a quarry launches a rock straight upward, and 2.4 s later it is rising at a rate of 10 m/s. assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.1 s after launch.

Let u = the vertical speed at launch, m/s

g = 9.8 m/s², acceleration due to gravity.

Part (a)

After 2.4 s, the vertical speed is 10 m/s.

Therefore

(10 m/s) = (v ft/s) - (9.8 m/s²) * (2.4 s)

v = 10 + 9.8*2.4 = 33.52 m/s

Answer: 33.52 m/s upward

Part (b)

After 5.1 s, the vertical speed is

(33.52 m/s) - (9.8 m/s²) * (5.1 s) = - 16.46 m/s.

This means that the rock reached the maximum height and began falling downward.

Answer: 16.46 m/s downward