The trajectory of a projectile, projected from the ground, is given by where and are measured in meter. The maximum height attained by the projectile will be. [2023]

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Solution

Q.
The trajectory of a 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. [2023]

1 $10 \sqrt{2}$ m

2 200 m

3 10 m

4 5 m

Ans.
(4)

$y = x - \frac{x^{2}}{20}$

$For maximum height, \frac{d y}{d x} = 0 \to 1 - \frac{2 x}{20} = 0$

$x = 10$

$So, y_{\max} = 10 - \frac{100}{20} = 5 m$

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