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This is a physics problem involving projectile motion. We need to find the speed of the rock just before it strikes the ground. Let’s work through the solution step by step:

### Given Information
- Initial velocity (\(v_0\)) = 8.00 m/s
- Angle of projection (\(\theta\)) = 37.0° above the horizontal
- Height of the building (\(h\)) = 10.0 m
- Acceleration due to gravity (\(g\)) = 9.81 m/s²

### Step 1: Resolve the Initial Velocity into Components
The initial velocity has horizontal and vertical components:
- \(v_{0x} = v_0 \cos(\theta) = 8.00 \cos(37.0°)\)
- \(v_{0y} = v_0 \sin(\theta) = 8.00 \sin(37.0°)\)

Calculating these:
- \(v_{0x} = 8.00 \cos(37.0°) \approx 8.00 \times 0.7986 = 6.39 \text{ m/s}\)
- \(v_{0y} = 8.00 \sin(37.0°) \approx 8.00 \times 0.6018 = 4.81 \text{ m/s}\)

### Step 2: Find the Final Vertical Velocity
The final vertical velocity (\(v_y\)) just before it hits the ground can be found using the kinematic equation:
\[
v_y^2 = v_{0y}^2 + 2g h
\]
\[
v_y^2 = (4.81 \, \text{m/s})^2 + 2(9.81 \, \text{m/s}^2)(10.0 \, \text{m})
\]
\[
v_y^2 = 23.1361 + 196.2 = 219.3361
\]
\[
v_y = \sqrt{219.3361} \approx 14.81 \, \text{m/s}
\]

### Step 3: Calculate the Magnitude of the Final Velocity
The horizontal velocity (\(v_x\)) remains unchanged because there is no horizontal acceleration:
\[
v_x = 6.39 \, \text{m/s}
\]

The total speed (\(v\)) of the rock just before impact is given by:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
\[
v = \sqrt{(6.39 \, \text{m/s})^2 + (14.81 \, \text{m/s})^2}
\]
\[
v = \sqrt{40.8321 + 219.3361}
\]
\[
v = \sqrt{260.1682}
\]
\[
v \approx 16.1 \, \text{m/s}
\]

### Conclusion
The correct answer is (d) 16.1 m/s.

Do you have any questions or would you like further details?

Here are some related questions:
1. How would the speed change if the angle of projection were different?
2. What if the height of the building was doubled; how would that affect the final speed?
3. How do you calculate the time it takes for the rock to reach the ground?
4. What would happen if there were air resistance? How would that affect the speed calculation?
5. How does the initial velocity affect the horizontal range of the projectile?

**Tip:** Always break down the velocity into its horizontal and vertical components in projectile motion problems for clarity.

A small rock is projected from the roof of a 10.0 m tall building. The initial velocity of the rock has a magnitude of 8.00 m/s and a direction 37.0° above the horizontal. What is the speed of the rock just before it strikes the ground?

Solve the physics problem where a rock is projected from a 10.0 m tall building with an initial velocity of 8.00 m/s at a 37.0° angle above the horizontal. Learn how to calculate the final speed before the rock strikes the ground using kinematic equations and trigonometry. Suitable for high school physics students.

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