The equation of a trajectory of a projectile isy = 8.1 - 422. The maximum height of the projectile is
(A) 8 m
(B) 6 m
(C) 2 m
(D) 4 m
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Answered by
1
For maximum height:
dxdy=0⇒dxd(10x−95x2)=0
⇒10−910x=0
⇒x=9 m
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0
Answer:
The answer is a option A
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