The trajectory of a projectile in vertical plane in y = ax - bx^ 2...

The trajectory of a projectile in vertical plane in

y = ax - bx^{2},

where

a

and

b

are constant and

x

and

y

are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal is: The trajectory of a projectile in vertical plane in y = ax - bx^ 2 , where a and b are constant and x and y are

The trajectory of a projectile in vertical plane in y = ax - bx^ 2 , where a and b are constant and x and y are

Options

A $\dfrac{a^{2}}{4b}, \tan^{-1}( b )$
B $\dfrac{a^{2}}{b}, \tan^{-1}( 2b )$
C $\dfrac{a^{2}}{4b}, \tan^{-1}( a )$
D $\dfrac{{2a}^{2}}{b}, \tan^{-1}( a )$

Correct answer

C. $\dfrac{a^{2}}{4b}, \tan^{-1}( a )$

Step-by-step solution

Let the angle of projection be

\text{\theta}

y = ax - {bx}^{2}

To get maxima,

\dfrac{\text{dy}}{\text{dx}} = 0

\dfrac{\text{dy}}{\text{dx}} = \text{a} - \text{b} . \text{2x} = 0

⇒ \text{x} = \dfrac{\text{a}}{\text{2b}}

{y}_\max = \dfrac{a^{2}}{2 b} - \dfrac{{ba}^{2}}{4 b^{2}} = \dfrac{a^{2}}{\text{4b}}

Angle of projections can be obtained by slope \left(\dfrac{\text{dy}}{\text{dx}}\right)

at

\text{x} = 0

<br /><br />

⇒

Slope at

\text{x} = 0 is \text{a}

<br /><br />

⇒ \text{tan} \theta = \text{a}

<br /><br />

⇒ \theta = \tan^{-1} \left(\text{a}\right)

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