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
$F = mleft(rac{du}{dt}
ight)$
$F = mg$
$F = mleft(rac{dy}{dt}
ight)$
$F = -mg$
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2023