If time of flight of a projectile is 10 secon | vedclass

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Question

(a)$T = \frac{{2u\sin 	heta }}{g} = 10\sec \Rightarrow u\sin 	heta = 50\;m/s$

 $H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} = \frac{{{{(u\sin

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$125 $

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(a)$T = \frac{{2u\sin 	heta }}{g} = 10\sec \Rightarrow u\sin 	heta = 50\;m/s$

 $H = \frac{{{u^2}{{\sin }^2}	heta }}{{2g}} = \frac{{{{(u\sin 	heta )}^2}}}{{2g}} = \frac{{50 	imes 50}}{{2 	imes 10}} = 125m$

Column $-I$ $R/H_{max}$	Column $-II$ Angle of projection $\theta $
$A.$ $1$	$1.$ ${60^o}$
$B.$ $4$	$2.$ ${30^o}$
$C.$ $4\sqrt 3$	$3.$ ${45^o}$
$D.$ $\frac {4}{\sqrt 3}$	$4.$ $tan^{-1}\,4\,=\,{76^o}$

If time of flight of a projectile is $10$ seconds. Range is $500$ meters. The maximum height attained by it will be ......... $m$

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