A player throws a ball that reaches to the an | vedclass

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Question

$\mathrm{T}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}=4 \mathrm{s}$

$	herefore u \sin 	heta=2 g$

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2}

Vedclass · Accepted Answer

$21.1$

Vedclass · Answer

$\mathrm{T}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}=4 \mathrm{s}$

$	herefore u \sin 	heta=2 g$

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}=\frac{4 \mathrm{g}^{2}}{2 \mathrm{g}}=2 \mathrm{g}$

$2 	imes 9.8=19.6 \mathrm{m}$

Height of the ball above the ground $=19.6+1.5$

$=21.1 \mathrm{m}$

A player throws a ball that reaches to the another player in $4\,s$. If the height of each player is $1.5\,m$, the maximum height attained by the ball from the ground level is .......... $m$

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