A dynamite blast at a quarry launches a rock straight upward, and 1.5 s later it is rising at a rate of 19 m/s. Assuming air resistance has no effect on the rock, calculate its speed (a) at launch and (b)4.6 s after launch.

V = Vo - gt = +19 m/s at t = 1.5

(a) Solve for Vo
Vo = 19 + 9.8*1.5 = 33.7 m/s

(b) Plug in t = 4.6 in the first formula and solve for V, which will be negative.

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To solve this problem, we can use the principles of physics, specifically the equations of motion. There are a few steps we need to follow to find the speed of the rock.

Step 1: Identify the known values:
We are given the following information:
- The time at which the rock is rising at a rate of 19 m/s is 1.5 seconds after launch.
- The acceleration due to gravity (g) is approximately 9.8 m/s².

Step 2: Determine the initial velocity (v₀):
The initial velocity of the rock at launch is what we need to find.

Step 3: Use the kinematic equation to calculate the initial velocity (v₀):
We can use the equation: v = v₀ + gt, where v is the final velocity, v₀ is the initial velocity, g is the acceleration due to gravity, and t is the time.

(a) Solving for the initial velocity at launch (v₀):
Given that the final velocity (v) is 19 m/s and the time (t) is 1.5 s, we can rearrange the equation to solve for v₀:
19 m/s = v₀ + (9.8 m/s²)(1.5 s)
19 m/s - (9.8 m/s²)(1.5 s) = v₀
v₀ ≈ 3.65 m/s

So, the initial velocity of the rock at launch is approximately 3.65 m/s.

(b) Calculating the speed 4.6 seconds after launch:
To calculate the speed of the rock at 4.6 seconds after launch, we will use the same equation: v = v₀ + gt.

v = (3.65 m/s) + (9.8 m/s²)(4.6 s)
v ≈ 48.7 m/s

Therefore, the speed of the rock approximately 4.6 seconds after launch is 48.7 m/s.