A particle of mass 2 m is projected at an angle of 45^ ^ with the...

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 $v_{y}=u_{y}-g t$ Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second part will become $v_y^{\prime}=20 m / s$ 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} / \mathrms^{2}\mathrm{h}_{2}=\frac{\mathrm{v}_{\mathrmy^{\prime 2}}{2 \mathrm{ g}}=\frac{(20)^{2}}{2 \times 10}=20 \mathrm{ m}\mathrm{H}=20+15=35 \mathrm{ m}