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Q.

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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a

4 m

b

1 m

c

2 m

d

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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