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A particle is projected upwards . the times corresponding to height h while ascending and while descending are 4s and 6s respectively . Find the velocity of projection

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3 years ago

```							A particle is projected upwards.Height = h time of ascent = t1time of descent =t2.a=-glet u be the initial velocity.u=?By using second equations of motions,S=ut+1/2 at²h=ut-1/2gt²2h=2ut-gt²gt²-2ut-+2h=0This is quadratic equation which corresponds in t which shows that for a given h there are two values of t [t1 and t2]By comparing the above equation with quadratic equation,ax²-bx+c=0we have :sum of roots =t1+t2= -b/a= -(-2u)/g=2u/gProduct of roots=t1t2=c/a=2h/g∴ sum of roots, t1+t2=2u/g∴u=g(t1+t2)/2∴Velocity of projection is g(t1+t2)/2hence velocity of projection in question v = g(4+6)/2v = 5g
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3 years ago
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