InterviewSolution
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InterviewSolution
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A ball is projected from the ground with velocity v such that its ranege is maximum. |
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Answer» Correct Answer - `ararrp;brarrq;crarrr;drarrs` The range of projectile is maximum. `therefore theta=45^(@)` and `H=(v^(2)sin^(2)45^(@))/(2g)=(v^(2))/(4g)` (a) Using, `v_(y)^(2)-u_(y)^(2)=2a_(y)Y` `v_(y)^(2)-(v^(2))/(2)=2(-g)(H)/(2)` `=-g((v^(2))/(4g))=-(v^(2))/(4)` `therefore v_(y)^(2)=(v^(2))/(4) rArr v_(y)=(v)/(2)` `therefore (a)rarr(p)` (b) At the maximum height, the ball has only horizontal velocity. `therefore v_(x)=(v)/sqrt(2)` `therefore (b)rarr(q)` (c) Horizontal velocity remains the same but the vertical velocity gets reversed. `therefore Deltav_(y)=(v)/sqrt(2)-(-(v)/sqrt(2))=(2v)/sqrt(2)=vsqrt(2)` `therefore (c)rarr(r)` (d) `vecv_(av)=(1)/(2)(vecv_(1)+vec_(f))` `=(1)/(2)[(v)/sqrt(2)(hati+hatj)+(v)/sqrt(2)hati]` `=(1)/(2)[(2v)/sqrt(2)hati+(v)/sqrt(2)hatj]` `therefore |vecv_(av)|=(1)/(2)sqrt(((4v^(2))/(2)+(v^(2))/(2)))=(v)/(2)sqrt((5)/(2))` `therefore (d) rarr(s)` |
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