To find the speed of the ball when it has fallen 20 m from the top of the tower, we can use the principle of conservation of mechanical energy. The total mechanical energy at the initial point (when the ball is thrown) must be equal to the total mechanical energy at the point when the ball has fallen 20 m.
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Identify initial and final states:
- Initial height \( h_i = 70 , \text{m} \)
- Final height \( h_f = 70 , \text{m} - 20 , \text{m} = 50 , \text{m} \)
- Initial velocity \( v_i = 3 , \text{m/s} \)
- Final velocity \( v_f = ? \)
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Calculate initial mechanical energy: The mechanical energy consists of kinetic energy (KE) and potential energy (PE). The initial mechanical energy \( E_i \) is given by:
\[ E_i = KE_i + PE_i \]
The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (3^2) = \frac{9}{2} m \]
The initial potential energy \( PE_i \) is:
\[ PE_i = mgh_i = mg(70) \]
Therefore,
\[ E_i = \frac{9}{2} m + mg(70) \]
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Calculate final mechanical energy: The final mechanical energy \( E_f \) is also given by:
\[ E_f = KE_f + PE_f \]
The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} m v_f^2 \]
The final potential energy \( PE_f \) is:
\[ PE_f = mgh_f = mg(50) \]
Therefore,
\[ E_f = \frac{1}{2} m v_f^2 + mg(50) \]
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Set initial mechanical energy equal to final mechanical energy:
\[ E_i = E_f \] \[ \frac{9}{2} m + mg(70) = \frac{1}{2} m v_f^2 + mg(50) \]
We can cancel \( m \) from both sides since it does not equal zero, and we can simplify to:
\[ \frac{9}{2} + g(70) = \frac{1}{2} v_f^2 + g(50) \]
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Substituting \( g \):
Taking \( g = 9.81 , \text{m/s}^2 \):
\[ \frac{9}{2} + 9.81(70) = \frac{1}{2} v_f^2 + 9.81(50) \]
Calculate \( 9.81(70) \) and \( 9.81(50) \):
\[ 9.81(70) = 686.7 \quad \text{and} \quad 9.81(50) = 490.5 \]
Substitute these values back into the equation:
\[ \frac{9}{2} + 686.7 = \frac{1}{2} v_f^2 + 490.5 \] \[ 4.5 + 686.7 = \frac{1}{2} v_f^2 + 490.5 \] \[ 691.2 = \frac{1}{2} v_f^2 + 490.5 \]
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Isolating \( v_f^2 \):
Subtract \( 490.5 \) from both sides:
\[ 691.2 - 490.5 = \frac{1}{2} v_f^2 \] \[ 200.7 = \frac{1}{2} v_f^2 \] \[ v_f^2 = 2 \times 200.7 = 401.4 \]
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Taking the square root:
\[ v_f = \sqrt{401.4} \approx 20.03 , \text{m/s} \]
Thus, the speed of the ball when it has fallen 20 m from the top of the tower is approximately 20.03 m/s.