A boy throws a ball vertically upward in air in such a manner that when the ball is in its maximum height he throws another ball. If the balls are thrown after the time difference of 1, sec then what will be the height attained by them ?

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19.6m

4.9 m

2.45m

9.8m

answer is C.

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Detailed Solution

Each ball rises for one second. Also, the velocity is reduced to zero in one second. Hence
$v = u - gt or 0 = u - 9 \cdot 8 \times 1$
or u = 9.8m/s
$∴$ height attained $h = 9 \cdot 8 \times 1 - \frac{1}{2} \times 9 \cdot 8 \times (1)^{2}$
= 4.9 m

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