Trajectory of projectile, projected from ground is given by y=x-frac{x^2}{20} . Where x and y are me

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3-2.Motion in Plane

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The trajectory of projectile, projected from the ground is given by $y=x-\frac{x^2}{20}$. Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be $...........\,m$

A$5$

B$10 \sqrt{2}$

C$200$

D$10$

(JEE MAIN-2023)

Solution

$y=x-\frac{x^2}{20}$
For maximum height,
$\frac{ dy }{ dx }=0 \Rightarrow 1-\frac{2 x }{20}=0$
$x=10$
So, $y _{\max }=10-\frac{100}{20}=5\,m$

Standard 11

Physics

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