The motion of a particle of mass m is described by the relation, $y =

The motion of a particle of mass m is described by the relation, $y = ut – \frac{1}{2}gt^2$, where $u$ is the initial velocity of the particle. The force acting on the particle is

$F = mleft( rac{du}{dt} ight)$

$F = mg$

$F = mleft( rac{dy}{dt} ight)$

$F = -mg$

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Answer is Wrong!

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UPSC NDA-2 – 2023

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The force acting on the particle is $F = -mg$.

The given equation of motion, $y = ut – \frac{1}{2}gt^2$, describes uniformly accelerated motion under gravity. By taking the second derivative of the displacement ($y$) with respect to time ($t$), we find the acceleration. The first derivative gives velocity: $\frac{dy}{dt} = u – gt$. The second derivative gives acceleration: $\frac{d^2y}{dt^2} = -g$. According to Newton’s second law, the force is given by $F = ma$. Since the acceleration is $a_y = -g$, the force is $F_y = m(-g) = -mg$. The negative sign indicates the force is acting in the negative y direction, which corresponds to the downward direction (gravity) if the initial upward direction was taken as positive y.

The equation $y = ut – \frac{1}{2}gt^2$ is the standard kinematic equation for vertical displacement under constant gravitational acceleration ($g$), with initial velocity $u$. The force responsible for this motion is the gravitational force, which is $mg$ acting downwards. Assuming the upward direction as positive y, the downward force is represented as $-mg$.

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