A ball thrown by one player reaches the other | vedclass

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Question

$\mathrm{t}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}} \quad 	ext { or } \quad 2=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}$

$	herefore \mathrm

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$\mathrm{t}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}} \quad 	ext { or } \quad 2=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}$

$	herefore \mathrm{usin} 	heta=\mathrm{g}$

$	herefore \quad \mathrm{h}_{\mathrm{m}}=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}=\frac{\mathrm{g}^{2}}{2 \mathrm{g}}=\frac{\mathrm{g}}{2}=5 \mathrm{m}$

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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