A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.3 s later it is rising at a speed of 18 m/s. Assuming air resistance has no effect on the rock, calculate its speed at (a) at launch and (b) 5.4 s after the launch.

v₁=v₀-gt₁

v₀=v₁+ gt₁=18+9.8•5.4= 40.54 m/s
At the top point v=0
0 =v₀-gt
t = v₀/g=40.54/9.8 =4.14 s.
t₃=t₂- t=5.4 – 4.14 = 1.26 s
v₂=gt₃ = 9.8•1.26 = 12.38 m/s (directed dowmward)

To calculate the speed of the rock at different times, we can use the equations of motion.

(a) To find the speed at launch, we need to consider the initial velocity of the rock. Since the rock starts from rest, the initial velocity is 0 m/s.

(b) To find the speed 5.4 seconds after the launch, we can use the equation:
v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
u = 0 m/s (initial velocity)
t = 5.4 s (time)

First, let's calculate the acceleration. Since the only force acting on the rock is gravity, we can use the formula:
a = g = 9.8 m/s^2

Now, we can calculate the final velocity:
v = u + at
v = 0 + (9.8)(5.4)
v ≈ 52.92 m/s

So, the speed of the rock 5.4 seconds after the launch is approximately 52.92 m/s.

To summarize:
(a) The speed at launch is 0 m/s.
(b) The speed 5.4 seconds after the launch is approximately 52.92 m/s.