15 Sy A dynamite blast at a quarry launches a chunk of
rock straight upward, and 2.0 s later it is rising at a speed
of . Assuming air resistance has no effect on the
rock, calculate its speed:
at launch;
5,0 s after launch.

Ask by Garza Deleon. in Italy

Jan 24,2025

Question

15 Sy A dynamite blast at a quarry launches a chunk of
rock straight upward, and 2.0 s later it is rising at a speed
of . Assuming air resistance has no effect on the
rock, calculate its speed:
at launch;
5,0 s after launch.

Ask by Garza Deleon. in Italy

Jan 24,2025

UpStudy AI · Accepted Answer

To solve this problem, we can use the kinematic equations for uniformly accelerated motion. We'll assume upward is the positive direction and take the acceleration due to gravity $$ g = 9.8 \, \text{m/s}^2 $$ (acting downward).

### 1. **Speed at Launch**

Let $$ v_0 $$ be the initial velocity (speed at launch).

The velocity of the rock after time $$ t $$ is given by:
$$
v = v_0 - g t
$$

We know that $$ v = 15 \, \text{m/s} $$ at $$ t = 2.0 \, \text{s} $$. Plugging in the values:
$$
15 \, \text{m/s} = v_0 - (9.8 \, \text{m/s}^2)(2.0 \, \text{s})
$$
$$
15 \, \text{m/s} = v_0 - 19.6 \, \text{m/s}
$$
$$
v_0 = 15 \, \text{m/s} + 19.6 \, \text{m/s} = 34.6 \, \text{m/s}
$$

**Initial Speed at Launch:** $$ \boxed{34.6 \, \text{m/s} \text{ upward}} $$

### 2. **Speed 5.0 Seconds After Launch**

Using the same equation for $$ v $$ at $$ t = 5.0 \, \text{s} $$:
$$
v = 34.6 \, \text{m/s} - (9.8 \, \text{m/s}^2)(5.0 \, \text{s})
$$
$$
v = 34.6 \, \text{m/s} - 49 \, \text{m/s} = -14.4 \, \text{m/s}
$$

The negative sign indicates the rock is moving **downward** at this time.

**Speed at 5.0 Seconds After Launch:** $$ \boxed{14.4 \, \text{m/s} \text{ downward}} $$

### **Summary**
- **At Launch:** $$ 34.6 \, \text{m/s} $$ upward
- **After 5.0 Seconds:** $$ 14.4 \, \text{m/s} $$ downward
At launch, the rock was moving upward at 34.6 m/s. Five seconds after launch, it was moving downward at 14.4 m/s.

15 Sy A dynamite blast at a quarry launches a chunk of
rock straight upward, and 2.0 s later it is rising at a speed
of . Assuming air resistance has no effect on the
rock, calculate its speed:
at launch;
5,0 s after launch.

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Latest Physics Questions

15 Sy A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of . Assuming air resistance has no effect on the rock, calculate its speed: at launch; 5,0 s after launch.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Physics Questions

15 Sy A dynamite blast at a quarry launches a chunk of
rock straight upward, and 2.0 s later it is rising at a speed
of . Assuming air resistance has no effect on the
rock, calculate its speed:
at launch;
5,0 s after launch.