 Projection Under uniform Acceleration
Question

# A particle is projected at an angle of ${45}^{\mathrm{o}}$ . After 1 sec, it breaks into two equal parts. One part stops and the other part attains the height of 20 m after the breaking of the particle. Find the velocity of projection ( g = 10 $\mathrm{m}/{\mathrm{s}}^{2}$ )

Moderate
Solution

## $\frac{{\mathrm{V}}_{\mathrm{y}}^{2}}{2\mathrm{g}}=20\mathrm{m}$$⇒{\mathrm{V}}_{\mathrm{y}}=20\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$ = vertical velocity of the second part.From conservation of momentum  $\mathrm{m}×0+{\mathrm{mV}}_{\mathrm{y}}=\left(2\mathrm{m}\right){\mathrm{V}}_{{\mathrm{y}}_{\mathrm{o}}}$${\mathrm{V}}_{{\mathrm{y}}_{0}}$  vertical velocity of particle after one second before breaking of it= $\frac{{\mathrm{V}}_{\mathrm{y}}}{2}=\frac{20}{2}=10\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$But  ${V}_{{y}_{0}}=10={u}_{y}-gt$ $⇒{\mathrm{u}}_{\mathrm{y}}=10+10×1=20\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$Now, ${u}_{y}=u\mathrm{sin}{45}^{o}$$⇒\mathrm{u}=\frac{{\mathrm{u}}_{\mathrm{y}}}{\mathrm{sin}{45}^{\mathrm{o}}}=20\sqrt{2}\text{\hspace{0.17em}}\mathrm{m}/\mathrm{s}$

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