# a canon is to fire up to 500 m horizontally.what should be the angle of projection , if the shells are fired witha velocity of 100 m/s?(g=10 m/s2)​

Dear Student,

$R=\frac{{u}^{2}\mathrm{sin}2\theta }{g}=\frac{100×100×sin2\theta }{10}\phantom{\rule{0ex}{0ex}}sin2\theta =\frac{R×10}{100×100}\frac{500×10}{100×100}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}2\theta ={30}^{\circ }\phantom{\rule{0ex}{0ex}}\theta ={15}^{\circ }$

Regards,
Manoj Singh

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