InterviewSolution
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A particle is projected with speed `v_(0)` at an angle `theta(theta ne 90^(@))` with horizontal and it bounce at same angle with horizontal. If average velocity of journey is `0.8 v_(0)` where `v_(0)` is average velocity of first projectile then `alpha` is |
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Answer» For first projectile =((2u_(x_(1))u_(y_(1)))/g+(2u_(x_(2))u_(y_(2)))/g+.......(2u_(x_(n))u_(y_(n)))/g)/((2u_(y_(1)))/g+....(2u_(y_(n)))/g)` `U_(x)[(1+1/(alpha^(2))+1/(alpha^(4))+.....1/(alpha^(2n)))/(1+1/(alpha)+1/(alpha^(2))+.....+1/(alpha^(n)))]=0.8 V_(0)` `(V_(0)[1/(1-1/(alpha^(2)))])/([1/(1-1/(alpha))])=0.8 v_(0)implies (alpha)/(1+alpha) =0.8 implies alpha=4` |
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