A dynamite blast at a quarry launches a rock straight upward, and 1.9 s later it is rising at a rate of 14 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b) 5.6 s after launch.

To solve this problem, we can use the kinematic equation relating the final velocity, initial velocity, time, and acceleration. The equation is:

v = u + at

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.

(a) To find the initial velocity at launch, we can use the given information that 1.9 seconds after launch, the rock is rising at a rate of 14 m/s.

Given:
Final velocity (v) = 14 m/s,
Time (t) = 1.9 s.

Using the equation, we can rearrange it to solve for the initial velocity:

v = u + at

Rearranging for u:

u = v - at

Substituting the given values:

u = 14 m/s - (9.8 m/s^2)(1.9 s)
u = 14 m/s - 18.62 m/s
u ≈ -4.62 m/s

Therefore, the initial velocity at launch is approximately -4.62 m/s. The negative sign indicates that the rock was launched upward.

(b) To find the speed of the rock 5.6 seconds after launch, we can use the same equation with the values provided.

Given:
Time (t) = 5.6 s.

Using the equation:

v = u + at

Substituting the given values:

v = -4.62 m/s + (9.8 m/s^2)(5.6 s)
v = -4.62 m/s + 54.88 m/s
v ≈ 50.26 m/s

Therefore, the speed of the rock 5.6 seconds after launch is approximately 50.26 m/s.