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A particle is projected up the inclined such that its component of velocity along the incline is `10 m//s`. Time of flight is `2` sec and maximum height above the incline is `5m`. Then velocity of projection will beA. `10 m//s`B. `10 sqrt(2) m//s`C. `5sqrt(5) m//s`D. none |
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Answer» Correct Answer - B Using the given data in the formulae for projection up the inclined plane. (`theta` is angle of projectile with inclined plane, `beta` is angle of inclined with horizontal) `T=2 sec =(2v sin theta)/(g cos beta)` `v cos theta=10.....(1)` `h=5 =(v sin theta)/2xx1` `v sin theta =10` from (i) and (ii) `v^(2)=200 ` `v=10sqrt(2) m//s` |
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