Q 11-03-102JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
The trajectory of projectile, projected from the ground is given by $y=x-\dfrac{x^2}{20}$, where $x$ and $y$ are measured in metre. The maximum height attained by the projectile will be
Answer: (C) $5\ \text{m}$
$\dfrac{dy}{dx}=1-\dfrac{x}{10}=0$ at $x=10$, giving $y_{max}=10-5=5$ m.
Solution by Sreeraj P, M.Sc Physics