The acceleration due to gravity on a planet is $1.96m/{{s}^{2}}.$ If it is safe to jump from a height of $3m$ on the earth, then corresponding height on the planet will be:
A.\[3m\]
B. $6m$
C. $9m$
D. $15m$

Hint: Acceleration due to gravity is the acceleration acting on the body due to the gravitational force. The kinematics (motion) of the body is clearly explained with the help of Newton's equation of motion. These equations are universally accepted. Every object which is performing motion obeys these equations.

Complete step by step answer:
As per the given data,
Acceleration due to gravity on the planet is $1.96m{{s}^{-2}}$
Safe height to jump on earth is $3m$
If the momentum of the body falling from the safe height of the earth is achieved by a body falling from a certain height on any planet. We can calculate the safe height on that planet.
So as per the given data safe height on the earth is $3m$.
From Newton $3^{rd}$ equation of motion, we know that.
${{v}^{2}}={{u}^{2}}+2as$
For free-falling body from rest, the equation can be written as,
${{v}^{2}}=2gs$
So the square velocity of the body falling from 3 m on earth will be,
$\begin{align}
  & {{v}^{2}}=2(9.8)3 \\
 & \Rightarrow {{v}^{2}}=58.8 \\
\end{align}$
To achieve the same moment on the other planet the body should fall with the same velocity as of earth. So, by putting the available information in the $3^{rd}$ equation of motion,
$\begin{align}
  & 58.8=2(1.96)s \\
 & \Rightarrow s=\dfrac{58.8}{2(1.96)} \\
 & \Rightarrow s=15m \\
\end{align}$
Thus the safe height for the planet having the value of $g=1.96m{{s}^{-2}}$ is $15m$.

So, the correct answer is “Option D”.

Note: When an object is free-falling the body has an acceleration the same as that of acceleration due to gravity acting on it. The value of g (acceleration due to gravity) depends on the gravitational force acting on the object. The value of g is not constant in the universe.