Physics - Kinematics Question with Solution | TestHub

Horizontal component of velocity ${\mathrm{v}}_x=10\cos 60^{\circ }=5\mathrm{m}/\mathrm{s}$ vertical component of velocity $v_y=10\cos 30^{\circ }=5\sqrt{3}\mathrm{m}/\mathrm{s}$ After $\mathrm{t}=1\sec$ Horizontal component of velocity ${\mathrm{v}}_x=5\mathrm{m}/\mathrm{s}$ Vertical component of velocity $v_y=|(5\sqrt{3}-10)|\mathrm{m}/\mathrm{s}=10-5\sqrt{3}$ Centripetal, acceleration ${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{v}}^2}{\mathrm{R}}$ $\Rightarrow \mathrm{R}=\frac{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}{{\mathrm{a}}_{\mathrm{n}}}=\frac{25+100+75-100\sqrt{3}}{10\cos \theta }$ From figure (using (i)) $\tan \theta =\frac{10-5\sqrt{3}}{5}=2-\sqrt{3}\Rightarrow \theta =15^{\circ }$ $\mathrm{R}=\frac{100(2-\sqrt{3})}{10\cos 15}=2.8\mathrm{m}$