A ball is thrown up with some speed. During up and then down motion, it crosses a point $A$ at t = 6 sec and t = 8 sec. The maximum height attained by ball is, (ball projected at t = 0 sec) (g = 10 $m / s^{2}$ )

245 m

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315 m

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145 m

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205 m

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Solution

The correct option is A 245 m
By symmetry, the ball must be at maximum height at t = 7 sec.
Therefore, the time taken to reach the maximium height is 7 s.
$h = \frac{1}{2} a t^{2} = \frac{1}{2} 10 \times 7^{2} = 245 m$

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