A particle is projected at an angle of $$45o$$ . 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 $$=$$ $$10$$ $$m/s2$$ $$)$$

Question

A particle is projected at an angle of $$45o$$ . 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 $$=$$ $$10$$ $$m/s2$$ $$)$$

Infinity Learn · Accepted Answer

$$Vy22g=20m$$⇒$$Vy=20$$ $$m/s$$ $$=$$ vertical velocity of the second part.From conservation of momentum  m×$$0+mVy=2mVyoVy0$$  vertical velocity of particle after one second before breaking of $$it=$$ $$Vy2=202=10$$ $$m/sBut$$  $$Vy0=10=uy$$−gt ⇒$$uy=10+10$$×$$1=20$$ $$m/sNow$$, $$uy=usin45o$$⇒$$u=uysin45o=202$$ $$m/s$$

A particle is projected at an angle of $45^{o}$ . 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 $m / s^{2}$ )

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A particle is projected at an angle of 45o . 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 m/s2 )

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A particle is projected at an angle of $45^{o}$ . 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 $m / s^{2}$ )