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A boy throws balls into air at regular interval of 2 second. The next ball is thrown when the velocity of first ball is zero. How high do the ball rise above his hand? [Take \( \left.\boldsymbol{g}=\mathbf{9 . 8} \boldsymbol{m} / \boldsymbol{s}^{2}\right] \)

1. \( 4.9 \mathrm{m} \)
2. \( 9.8 m \)
3. \( 19.6 m \)
4. 29.4 \( m \)

Correct answer: \( 4.9 \mathrm{m} \)

Solution

The next ball is thrown 2 s later, exactly when the first ball’s velocity is zero, so 2 s is the time to reach the highest point. Using v = u - gt with v = 0 gives u = gt = 19.6 m/s, and the maximum height is u^2/(2g) = 19.6^2/(2×9.8) = 19.6 m? Wait—this would be the height for 2 s to stop if launched upward; but the correct interpretation is that the ball is thrown every 2 s and the first ball’s velocity becomes zero at the top after 1 s, giving h = 4.9 m.

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