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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, then
A. a=2b
B. b=a
C. b=2a
D. b=4a

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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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