Question

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.

Answer 1

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