Question

A particle is thrown vertically upwards with a velocity of $4m{{s}^{-1}}$. The ratio of its acceleration after 1s and 2s of its motion is

Answer 1

Earth exerts a gravitational force on a particle in the downwards direction. When the height of a particle above the surface of the earth is very less than the radius of the earth, the gravitational force exerted by the earth on the particle is constant.
Formula used:
F = mg
F = ma

Complete answer:
Earth exerts a gravitational force on a particle in the downwards direction. When the height of a particle above the surface of the earth is very less than the radius of the earth, the gravitational force exerted by the earth on the particle is constant. The magnitude of the gravitational force is given as F = mg, where m is the mass of the particle and g is the acceleration due to gravity and it is constant.
From Newton’s second law of motion we know that F = ma, where a is the acceleration of the particle.
Therefore, in the case of the particle influenced by gravitational force, its acceleration is equal to g.
When the particle is thrown vertically, the particle will accelerate downwards due to the force of gravity exerted by the earth. The acceleration of the particle will be equal to –g (i.e. acceleration due to gravity).
Since g is constant near the surface of earth, the acceleration of the particle will be constant throughout its motion.
Therefore, the acceleration of the particle at time t=1s will be –g. At time t=2s, its acceleration will be –g.
Therefore, the ratio of the acceleration of the particle after 1s and 2s of its motion is $\dfrac{-g}{-g}=1$.

So, the correct answer is “Option C”.

Note:
In questions which involve the motion of a particle with a constant acceleration, we can use the various kinematic equations.
That are:
$v=u+at$
$2as={{v}^{2}}-{{u}^{2}}$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$.
Here, u, v, a and s are the initial velocity, final velocity, acceleration and displacement of the particle.