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A projectile is thrown with an initial velocity of `(a hati +b hatj) ms^(-1)`. If the range of the projectile is twice the maximum height reached by it, thenA. a=2bB. b=aC. b=2aD. b=4a |
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Answer» Correct Answer - C `vecv=ahati+6hatjrArru cos theta=a` and `usin theta=b` `tantheta=(b)/(a)` also `tan theta=(4)/(n)=(4)/(2)=2` `therefore (b)/(a)=2 rArr b=2a` |
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