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Question Number 209099 by Spillover last updated on 01/Jul/24

Answered by mr W last updated on 02/Jul/24

say particle 2 starts time T  later  than particle 1.     particle 2 at time t:  h=ut−((gt^2 )/2)  particle 1 at time t+T:  h=u(t+T)−((g(t+T)^2 )/2)  when both particles meet at height h,  u(t+T)−((g(t+T)^2 )/2)=ut−((gt^2 )/2)  ⇒t=((2u−gT)/(2g))  h=ut−((gt^2 )/2)=((2u−gT)/(2g))×(u−(g/2)×((2u−gT)/(2g)))     =((4u^2 −g^2 T^2 )/(8g))  ✓

sayparticle2startstimeTlaterthanparticle1.particle2attimet:h=utgt22particle1attimet+T:h=u(t+T)g(t+T)22whenbothparticlesmeetatheighth,u(t+T)g(t+T)22=utgt22t=2ugT2gh=utgt22=2ugT2g×(ug2×2ugT2g)=4u2g2T28g

Commented by Spillover last updated on 02/Jul/24

thank you

thankyou

Answered by Spillover last updated on 02/Jul/24

Answered by Spillover last updated on 02/Jul/24

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