A ball thrown by one player reaches the other in 2 sec. The maximum height attained by the ball above the point of projection will be about

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2.5 m

5 m

7.5 m

10 m

answer is B.

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Detailed Solution

$\begin{array}{l} t = \frac{2 usin θ}{g} or 2 = \frac{2 usin θ}{g} \\ ∴ usin θ = g \\ ∴ h_{m} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{g^{2}}{2 g} = \frac{g}{2} = 5 m . \end{array}$

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