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When we analyse the projectile motion from any accelerated frame `O` as `vecr_(o),vecu_(o)` and `veca_(o)` respectively, express the following terms, `vecr_(pO)+veca_(p)-vecr_(O),vecu_(pO)=vecu_(p)-vecu_(O)` and `veca_(pO)=veca_(p)-veca_(O)` where `P` stands for projectile. Then using the following kinematical equations of the projectile (For constant acceleration) relative to the accelerating frame ,we have `vecS_(pO)=vecu_(pO)t+1/2veca_(pO)t^(2),vecv_(pO)` `=vecu_(pO)+veca_(pO)t` and `v_(pO)^(2)=u_(pO)^(2)+2veca.vecs_pO` Using the above expressions, anwer the following question:A projectile has initial velocity `v_(0)` realtive to the large plate which is moving with a constant upward acceleration a. Refering to Q.1, velocity of the projectile relative to `B` ofter some time A. `ltv_(0) "at an angle" thetalttheta_(0)`B. `gtv_(0) "at an angle" thetagttheta_(0)`C. `gtv_(0) "at an angle" theta=theta_(0)`D. `v_(0) "at an angle" theta=theta_(0)` |
Answer» Correct Answer - D Here`B` is ground frame, w.r.t `B` projection velocity will be greater than `V_(0)`. Hence when `thetagttheta_(0)` projectile velocity will be greater than `V_(0)`. |
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