Question

If a particle takes $t$ second less and acquires the velocity of $v$ $m/s$ more in falling through the same distance on to the planets where the acceleration due to gravity are $2g$ and $8g$ respectively. Then,
A. $4gt$
B. $5gt$
C. $2gt$
D. $16gt$

Answer 1

Hint: T We can relate the quantities like displacement, velocity, and time provided if the acceleration is constant through some equation. The relationship between speed, gravity, and time has to be used here. Calculate the values of acceleration, time taken, and distance for both planets separately.

Complete step by step solution:
Given there are two planets with their acceleration due to gravities are $2g$ and $8g$. A particle acquires the velocity $v$ $m/s$ and it takes time $t$ second.
Calculate the values of acceleration, time taken, and distance for both the planets separately.
For the first planet,
The acceleration is given by,
$ \Rightarrow a = 2g$
Where acceleration $ = a$ ,
And, gravity $ = g$ .
Time taken is $T$
Calculate the distance.
$ \Rightarrow D = \dfrac{1}{2}a{T^2}$
Substitute the values
$ \Rightarrow D = \dfrac{1}{2}2g{T^2}$
Where $D$ is the distance, $g$ is the gravity, and $T$ is the time taken.
Cancel out the common terms.
$ \Rightarrow D = g{T^2}$
Calculate the velocity. Velocity is defined by the product of acceleration and time. That is,
$ \Rightarrow v = aT$
Substitute the acceleration value,
$ \Rightarrow v = 2gT$
For the second planet,
The acceleration is given by,
$ \Rightarrow a = 8g$
Where $a$ is the acceleration, $g$ is the gravity.
Time taken is $T - t$
Calculate the distance.
$ \Rightarrow D = \dfrac{1}{2}a{T^2}$
Substitute the values.
$ \Rightarrow D = \dfrac{1}{2}8g{\left( {T - t} \right)^2}$
Simplify the equation.
$ \Rightarrow D = 4g{\left( {T - t} \right)^2}$
Calculate the velocity.
$ \Rightarrow v = aT$
Substitute the acceleration value,
$ \Rightarrow v = 8g\left( {T - t} \right)$
Compare the acceleration of the first and second planets.
$ \Rightarrow 8g\left( {T - t} \right) = 2gt + V$
Simplify the given equation,
$ \Rightarrow 6gT - 8gt = V$
Compare the distances of the first and second planets.
$ \Rightarrow g{T^2} = 4g{\left( {T - t} \right)^2}$
With the equation, try to find the value of time.
$ \Rightarrow {T^2} = 4{\left( {T - t} \right)^2}$
To remove the square power, take the square root. That is,
$ \Rightarrow T = 2\left( {T - t} \right)$
Multiply the $2$inside the bracket terms.
$ \Rightarrow T = 2T - 2t$
Hence the value of $T$is,
$ \Rightarrow T = 2t$
Substitute the value of $T$in the equation of $6gT - 8gt = V$
$ \Rightarrow 6g2t - 8gt = V$
The value of velocity is,
$ \Rightarrow V = 4gt$
Therefore option $\left( A \right)$ is the correct option.

Note:
Every object has a mass and every object that exerts mass will always have a gravitational pull. The strength of the pull will always depend on the mass of the objects. Gravity is the reason that keeps the planets in the orbit around the sun and the moon. This force pulls a body into the center of the earth or some other physical entity having a mass.