The trajectory of a projectile in a vertical plane is: y=ax−bx2, where a and b are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is
A
a24b
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B
a28b
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C
a22b
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D
a22gb
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Solution
The correct option is Aa24b The equation of trajectory is, y=ax−bx2 For maximum height of projectile, y should be maximum. ⇒dydx=0 and d2ydx2<0 ⇒dydx=a−2bx=0 ⇒x=a2b d2ydx2=−2b<0 ∴ For x=a2b, y is maximum Substituting the value of x=a2b in trajectory equation, ⇒ymax=a(a2b)−b(a2b)2 ∴ymax=a24b