JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Kinematics

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.

(A)5 m✓ Correct
(B) $102 m$
(C)200 m
(D)10 m

Answer: (A)

Solution

To find the maximum height, differentiate the trajectory equation with respect to x and set it to zero:

\frac{d y}{d x} = 1 - \frac{2 x}{20} = 0

. Solving gives

x = 10

m. Substituting back:

y_{max} = 10 - \frac{100}{20} = 10 - 5 = 5

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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