JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Kinematics

The trajectory of a projectile in a vertical plane is

y = α x - β x^{2}

, where

α

and

β

are constants and

x & y

are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection

θ

and the maximum height attained H are respectively given by :

Answer: (B)

Solution

The trajectory of a projectile launched at

θ

with speed

u

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

Matching it with

y = α x - β x^{2}

gives

tan θ = α

, so

θ = tan^{- 1} α

. The top of the path is where

\frac{d y}{d x} = 0

, that is

α - 2 β x = 0

, so

x = \frac{α}{2 β}

. Substituting this back,

H = α \cdot \frac{α}{2 β} - β (\frac{α}{2 β})^{2} = \frac{α ^{2}}{2 β} - \frac{α ^{2}}{4 β} = \frac{α ^{2}}{4 β}

The angle of projection is

tan^{- 1} α

and the greatest height reached is

\frac{α ^{2}}{4 β}

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option