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:
-
Horizontal component \(v_{xf}\):
- Since there is no horizontal acceleration, \(v_{xf} = v_{x0} = 6.0 , \text{m/s}\).
-
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.
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