# 41

Dear Student,

$ifvelocityoftheprojectileisuthen\phantom{\rule{0ex}{0ex}}Range=9=\frac{{u}^{2}\mathrm{sin}\left(90\right)}{g}\phantom{\rule{0ex}{0ex}}oru=3\sqrt{10}m/s\phantom{\rule{0ex}{0ex}}andtimetoreachpole=\frac{6}{u\mathrm{cos}\left(45\right)}=\frac{2}{\sqrt{5}}s\phantom{\rule{0ex}{0ex}}soheightofthepole=u\mathrm{sin}\left(45\right)t-\frac{1}{2}g{t}^{2}=3\sqrt{10}\times \frac{1}{\sqrt{2}}\times \frac{2}{\sqrt{5}}-\frac{1}{2}\times 10\times {\left(\frac{2}{\sqrt{5}}\right)}^{2}=2m\phantom{\rule{0ex}{0ex}}$

Regard

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