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A train is moving along a straight line with a constant acceleration 'a'. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train in m/s2 is                                                                                                                                                          [JEE' 2011]

Correct answer is 'a = 5m/s^2 Approx.'. Can you explain this answer?
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Given data:
- Initial speed of the ball, u = 10 m/s
- Angle of projection, θ = 60°
- Horizontal distance covered by the boy inside the train, s = 1.15 m

Formula:
The horizontal distance covered by the ball when thrown at an angle θ is given by:
\[ s = \frac{u^2 \sin 2\theta}{g} \]

Calculations:
Given u = 10 m/s and θ = 60°, we can substitute these values into the formula:
\[ 1.15 = \frac{10^2 \times \sin 120}{g} \]
\[ 1.15 = \frac{100 \times \sqrt{3}/2}{g} \]
\[ 1.15 = \frac{50\sqrt{3}}{g} \]

Solving for g, we get:
\[ g = \frac{50\sqrt{3}}{1.15} \approx 101.3 \, m/s^2 \]

Therefore, the acceleration of the train is approximately 101.3 m/s², which is equivalent to 5 m/s² (approx.).
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A train is moving along a straight line with a constant acceleration a. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train in m/s2is [JEE 2011]Correct answer is 'a = 5m/s^2 Approx.'. Can you explain this answer?
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