To calculate the kinetic energy of the ball when it reaches a height of 1.5 meters, we first need to determine the potential energy at both the initial height (3.25 meters) and the height just before it falls to 1.5 meters.
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Calculate the initial potential energy (PE_initial) when the ball is at 3.25 meters:
\[ PE_{\text{initial}} = mgh = (0.20 , \text{kg}) \times (9.81 , \text{m/s}^2) \times (3.25 , \text{m}) \]
\[ PE_{\text{initial}} = 0.20 \times 9.81 \times 3.25 \approx 6.39 , \text{joules} \]
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Calculate the potential energy (PE_final) when the ball is at 1.5 meters:
\[ PE_{\text{final}} = mgh = (0.20 , \text{kg}) \times (9.81 , \text{m/s}^2) \times (1.5 , \text{m}) \]
\[ PE_{\text{final}} = 0.20 \times 9.81 \times 1.5 \approx 2.94 , \text{joules} \]
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Determine the kinetic energy (KE) when the ball is at 1.5 meters above the ground. The kinetic energy can be found using the conservation of energy principle, which states that the potential energy lost equals the kinetic energy gained:
\[ KE = PE_{\text{initial}} - PE_{\text{final}} \]
\[ KE = 6.39 , \text{joules} - 2.94 , \text{joules} \approx 3.45 , \text{joules} \]
Rounding this to the appropriate significant figures, we can state that the kinetic energy of the ball when it reaches 1.5 meters above the ground is approximately:
\[ \boxed{3.4} \text{ joules} \]
So the correct answer is C. 3.4 joules.
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