Assertion (A): The velocity of a projectile a | vedclass

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(D) The trajectory of a projectile is given by the equation:$y = f(x) = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$The slope of the trajectory

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$A$ is false but $R$ is true

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(D) The trajectory of a projectile is given by the equation:$y = f(x) = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$The slope of the trajectory at any point $(x, y)$ is given by the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = 	an 	heta - \frac{gx}{u^2 \cos^2 	heta} = \frac{v_y}{v_x}$This slope represents the ratio of the vertical component of velocity to the horizontal component of velocity,not the magnitude of the velocity itself. Therefore,Assertion $(A)$ is incorrect.Reason $(R)$ states that the velocity vector at any point is always along the tangent to the trajectory at that point. This is a fundamental property of motion in a plane,as the instantaneous velocity is defined as the rate of change of position,which is directed along the tangent to the path. Thus,Reason $(R)$ is correct.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.
Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.

Similar Questions

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Match the columns:

Column-$I$ $(R/H_{max})$ Column-$II$ (Angle of projection $\theta$)

$A. 1$ $1. 60^o$

$B. 4$ $2. 30^o$

$C. 4\sqrt{3}$ $3. 45^o$

$D. 4/\sqrt{3}$ $4. \tan^{-1}(4) = 76^o$

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Assertion $(A)$: The velocity of a projectile at a point on its trajectory is equal to the slope at that point.Reason $(R)$: The velocity vector at a point is always along the tangent to the trajectory at that point.