Question

# Q.33. A projectile is projected with certain velocity at an angle $\theta$ with the horizontal from the ground. If its time of flight is T then the magnitude of change in its velocity when body reaches to ground again is (g is acceleration due to gravity) (1) gT (2) $\frac{gT}{2}$ (3) 2gT (4) $\frac{{g}^{2}}{T}$

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Solution

## Dear Student, $whenbodytouchesthegrounditwillmakeangle\theta with\phantom{\rule{0ex}{0ex}}horizontal\phantom{\rule{0ex}{0ex}}soanglebetweeninitialvelocityandfinalvelocityis2\theta \phantom{\rule{0ex}{0ex}}hencedifferenceintheirvelocities\phantom{\rule{0ex}{0ex}}∆v=\sqrt{{u}^{2}+{u}^{2}-2{u}^{2}\mathrm{cos}\left(2\theta \right)}=2u\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}butaccordingtoquestion\phantom{\rule{0ex}{0ex}}timeofflight=T=\frac{2u\mathrm{sin}\left(\theta \right)}{g}\phantom{\rule{0ex}{0ex}}⇒2u\mathrm{sin}\left(\theta \right)=gT\phantom{\rule{0ex}{0ex}}hence\phantom{\rule{0ex}{0ex}}∆v=gT$ Regard

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