Asked by goatmanthealien
Assignment #1 UNIT I “Projectile Motion”
Your work must be clear to read. Clearly label all cases (a, b, c). Show all steps in GRASS format. GRASS means given/required/analysis/solution/statement. You can type your solutions and answers, but neat handwriting is acceptable. Please submit your assignment as ONE PDF file on D2L. I do not accept any assignments via email.
Problem: A bowling ball is rolled off the top of a cliff with an initial horizontal velocity of 6.0 m/s. If the cliff is 100 m above the ground, determine:
a) the ball's time of flight
b) the ball's range
c) the ball's final velocity just before it strikes the ground.
Your work must be clear to read. Clearly label all cases (a, b, c). Show all steps in GRASS format. GRASS means given/required/analysis/solution/statement. You can type your solutions and answers, but neat handwriting is acceptable. Please submit your assignment as ONE PDF file on D2L. I do not accept any assignments via email.
Problem: A bowling ball is rolled off the top of a cliff with an initial horizontal velocity of 6.0 m/s. If the cliff is 100 m above the ground, determine:
a) the ball's time of flight
b) the ball's range
c) the ball's final velocity just before it strikes the ground.
Answers
Answered by
goatmanthealien
For case C, is the answer 44.74 m/s?
Answered by
GPT-4o mini
To solve the problem of the bowling ball rolled off a cliff, we will use the GRASS format to clearly outline our approach step by step.
### **Case a: The ball's time of flight**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Height of the cliff (\(h\)) = 100 m
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Time of flight (\(t\))
**Analysis:**
- The time of flight can be determined using the vertical motion equation. The motion in the horizontal direction does not affect the time it takes to fall. We will use the equation of motion for free fall:
\[
h = \frac{1}{2} g t^2
\]
**Solution:**
- Rearranging for \(t\):
\[
t^2 = \frac{2h}{g}
\]
\[
t^2 = \frac{2 \times 100 \, \text{m}}{9.81 \, \text{m/s}^2} \approx \frac{200}{9.81} \approx 20.387
\]
\[
t \approx \sqrt{20.387} \approx 4.52 \, \text{s}
\]
**Statement:**
- The ball's time of flight is approximately \(4.52\) seconds.
---
### **Case b: The ball's range**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (calculated from part a)
**Required:**
- Range (\(R\))
**Analysis:**
- The range can be calculated using the formula for horizontal motion:
\[
R = v_{x0} \cdot t
\]
**Solution:**
- Substituting the known values:
\[
R = 6.0 \, \text{m/s} \cdot 4.52 \, \text{s} \approx 27.12 \, \text{m}
\]
**Statement:**
- The ball's range is approximately \(27.12\) meters.
---
### **Case c: The ball's final velocity just before it strikes the ground**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (from part a)
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Final velocity (\(v_f\))
**Analysis:**
- The final velocity just before impact can be found by determining both the horizontal and vertical components:
1. **Horizontal component \(v_{xf}\)**:
- Since there is no horizontal acceleration, \(v_{xf} = v_{x0} = 6.0 \, \text{m/s}\).
2. **Vertical component \(v_{yf}\)**:
- We use the formula for vertical motion to find the final vertical velocity:
\[
v_{yf} = g \cdot t = 9.81 \, \text{m/s}^2 \cdot 4.52 \, \text{s} \approx 44.4 \, \text{m/s}
\]
**Solution:**
- The final velocity can be determined using the Pythagorean theorem:
\[
v_f = \sqrt{v_{xf}^2 + v_{yf}^2} = \sqrt{(6.0 \, \text{m/s})^2 + (44.4 \, \text{m/s})^2}
\]
Calculating:
\[
v_f = \sqrt{36 + 1971.36} = \sqrt{2007.36} \approx 44.83 \, \text{m/s}
\]
**Statement:**
- The ball's final velocity just before it strikes the ground is approximately \(44.83\) m/s.
---
### Final Summary
In summary:
- a) The ball's time of flight is approximately **4.52 seconds**.
- b) The ball's range is approximately **27.12 meters**.
- c) The ball's final velocity just before it strikes the ground is approximately **44.83 m/s**.
Remember to format your document properly and submit it as a single PDF file via the designated platform. Good luck!
### **Case a: The ball's time of flight**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Height of the cliff (\(h\)) = 100 m
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Time of flight (\(t\))
**Analysis:**
- The time of flight can be determined using the vertical motion equation. The motion in the horizontal direction does not affect the time it takes to fall. We will use the equation of motion for free fall:
\[
h = \frac{1}{2} g t^2
\]
**Solution:**
- Rearranging for \(t\):
\[
t^2 = \frac{2h}{g}
\]
\[
t^2 = \frac{2 \times 100 \, \text{m}}{9.81 \, \text{m/s}^2} \approx \frac{200}{9.81} \approx 20.387
\]
\[
t \approx \sqrt{20.387} \approx 4.52 \, \text{s}
\]
**Statement:**
- The ball's time of flight is approximately \(4.52\) seconds.
---
### **Case b: The ball's range**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (calculated from part a)
**Required:**
- Range (\(R\))
**Analysis:**
- The range can be calculated using the formula for horizontal motion:
\[
R = v_{x0} \cdot t
\]
**Solution:**
- Substituting the known values:
\[
R = 6.0 \, \text{m/s} \cdot 4.52 \, \text{s} \approx 27.12 \, \text{m}
\]
**Statement:**
- The ball's range is approximately \(27.12\) meters.
---
### **Case c: The ball's final velocity just before it strikes the ground**
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (from part a)
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Final velocity (\(v_f\))
**Analysis:**
- The final velocity just before impact can be found by determining both the horizontal and vertical components:
1. **Horizontal component \(v_{xf}\)**:
- Since there is no horizontal acceleration, \(v_{xf} = v_{x0} = 6.0 \, \text{m/s}\).
2. **Vertical component \(v_{yf}\)**:
- We use the formula for vertical motion to find the final vertical velocity:
\[
v_{yf} = g \cdot t = 9.81 \, \text{m/s}^2 \cdot 4.52 \, \text{s} \approx 44.4 \, \text{m/s}
\]
**Solution:**
- The final velocity can be determined using the Pythagorean theorem:
\[
v_f = \sqrt{v_{xf}^2 + v_{yf}^2} = \sqrt{(6.0 \, \text{m/s})^2 + (44.4 \, \text{m/s})^2}
\]
Calculating:
\[
v_f = \sqrt{36 + 1971.36} = \sqrt{2007.36} \approx 44.83 \, \text{m/s}
\]
**Statement:**
- The ball's final velocity just before it strikes the ground is approximately \(44.83\) m/s.
---
### Final Summary
In summary:
- a) The ball's time of flight is approximately **4.52 seconds**.
- b) The ball's range is approximately **27.12 meters**.
- c) The ball's final velocity just before it strikes the ground is approximately **44.83 m/s**.
Remember to format your document properly and submit it as a single PDF file via the designated platform. Good luck!
Answered by
GPT-4o mini
Let's re-evaluate the final velocity calculation in Case C to verify the result. We need to ensure all steps are clearly outlined and accurate.
### Case C: The Ball's Final Velocity Just Before It Strikes the Ground
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (from part a)
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Final velocity (\(v_f\))
**Analysis:**
1. **Horizontal component (\(v_{xf}\))**:
- Since there is no horizontal acceleration, the horizontal component of velocity remains:
\[
v_{xf} = v_{x0} = 6.0 \, \text{m/s}
\]
2. **Vertical component (\(v_{yf}\))**:
- The vertical component of the final velocity can be calculated using:
\[
v_{yf} = g \cdot t
\]
\[
v_{yf} = 9.81 \, \text{m/s}^2 \cdot 4.52 \, \text{s} \approx 44.4 \, \text{m/s}
\]
**Solution:**
Now we combine both components to find the resulting magnitude of the final velocity using the Pythagorean theorem:
\[
v_f = \sqrt{v_{xf}^2 + v_{yf}^2}
\]
Substituting the values we found:
\[
v_f = \sqrt{(6.0 \, \text{m/s})^2 + (44.4 \, \text{m/s})^2}
\]
Calculating each piece:
\[
v_f = \sqrt{36 + 1971.36} = \sqrt{2007.36}
\]
Now, calculating the square root:
\[
v_f \approx 44.83 \, \text{m/s}
\]
### Conclusion:
The final velocity just before it strikes the ground is indeed approximately **44.83 m/s**, not **44.74 m/s**.
If you have further context or a specific step where you believe the discrepancy may have arisen, feel free to share, and we can go through it together!
### Case C: The Ball's Final Velocity Just Before It Strikes the Ground
**Given:**
- Initial horizontal velocity (\(v_{x0}\)) = 6.0 m/s
- Time of flight (\(t\)) = 4.52 s (from part a)
- Acceleration due to gravity (\(g\)) = 9.81 m/s²
**Required:**
- Final velocity (\(v_f\))
**Analysis:**
1. **Horizontal component (\(v_{xf}\))**:
- Since there is no horizontal acceleration, the horizontal component of velocity remains:
\[
v_{xf} = v_{x0} = 6.0 \, \text{m/s}
\]
2. **Vertical component (\(v_{yf}\))**:
- The vertical component of the final velocity can be calculated using:
\[
v_{yf} = g \cdot t
\]
\[
v_{yf} = 9.81 \, \text{m/s}^2 \cdot 4.52 \, \text{s} \approx 44.4 \, \text{m/s}
\]
**Solution:**
Now we combine both components to find the resulting magnitude of the final velocity using the Pythagorean theorem:
\[
v_f = \sqrt{v_{xf}^2 + v_{yf}^2}
\]
Substituting the values we found:
\[
v_f = \sqrt{(6.0 \, \text{m/s})^2 + (44.4 \, \text{m/s})^2}
\]
Calculating each piece:
\[
v_f = \sqrt{36 + 1971.36} = \sqrt{2007.36}
\]
Now, calculating the square root:
\[
v_f \approx 44.83 \, \text{m/s}
\]
### Conclusion:
The final velocity just before it strikes the ground is indeed approximately **44.83 m/s**, not **44.74 m/s**.
If you have further context or a specific step where you believe the discrepancy may have arisen, feel free to share, and we can go through it together!