A ball thrown by one player reaches the other | vedclass

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(c) $\frac{{2u\sin 	heta }}{g} = 2\sec $ $&rArr;$ $u\sin 	heta = 10$

 $	herefore H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} = \frac{{1

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$5 $

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(c) $\frac{{2u\sin 	heta }}{g} = 2\sec $ $&rArr;$ $u\sin 	heta = 10$

 $	herefore H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} = \frac{{100}}{{2g}} = 5m$

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 ....... $m$

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