Questions

# 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 $\mathrm{m}/{\mathrm{s}}^{2}$ )

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a
20 m/s
b
202m/s
c
102m/s
d
15 m/s

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detailed solution

Correct option is B

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

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