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

Vedclass · Accepted Answer

$125 $

Vedclass · Answer

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

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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