Physics - Center of Mass Question with Solution | TestHub

Given : Initial velocity $u_0=20\sqrt{2}\mathrm{m}/\mathrm{s};$ angle of projection $\theta =45^{\circ }$ Therefore horizontal and vertical components of initial velocity are ${\mathrm{u}}_{\mathrm{x}}=20\sqrt{2}\cos 45^{\circ }=20\mathrm{m}/\mathrm{s}$ and ${\mathrm{u}}_{\mathrm{y}}=20\sqrt{2}\sin 45^{\circ }=20\mathrm{m}/\mathrm{s}$ After $1\mathrm{s}$, horizontal component remains unchanged while the vertical component becomes $v_y=u_y-gt$ Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second part will become $v_y^{\prime }=20m/s$ Therefore, maximum height attained by the second part will be $\mathrm{H}={\mathrm{h}}_1+{\mathrm{h}}_2$ Using, $\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ $\Rightarrow {\mathrm{h}}_1=(20\times 1)-\frac{1}{2}\times 10\times (1{)}^2=15\mathrm{m}$ $\mathrm{a}=\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$ ${\mathrm{h}}_2=\frac{{\mathrm{v}}_{\mathrm{y}}^{\prime 2}}{2\mathrm{g}}=\frac{(20{)}^2}{2\times 10}=20\mathrm{m}$ $\mathrm{H}=20+15=35\mathrm{m}$