The equation of trajectory of a projectile is given as y=2x−x22. The maximum height of projectile is (Symbols have usual meanings and SI unit)

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4 m

1 m

2 m

10 m

answer is C.

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Detailed Solution

y=2x−xh2 dydx=0⇒ dydx=2−2x2=0When y=hmax b ⇒x=2m∴hmax=2×2−2×22 =2m

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