Question

An object is falling freely under a gravitational force. Its velocity after travelling a distance $h$ is $V$. If $V$ depends upon gravitational acceleration $g$ and the distance, then prove with dimensional analysis that $V=k\sqrt{gh}$ where $k$ be a constant.

Answer 1

Hint : Velocity is given as the rate of change of displacement. And the acceleration is given as the rate of variation of the velocity. The displacement is the shortest distance covered by the particle. These all may help you to solve this question.

Complete answer:
Here the velocity is given by the equation,
$V=\dfrac{\text{displacement}}{\text{time}}$
The velocity is having the dimension,
$\left| V \right|=L{{T}^{-1}}$
Acceleration is the rate of change of velocity.
$a=\dfrac{\text{velocity}}{\text{time}}$
Therefore it will be having dimension as,
$\left| a \right|=L{{T}^{-2}}$
Displacement is the shortest distance travelled. Therefore its dimension will be,
$\left| h \right|=L$
Therefore, we can write as per the equation,
$V=k\sqrt{gh}$
Where $k$has been mentioned as a constant.
The dimensional formula can be applied to the terms given in the equation,
$L{{T}^{-1}}={{\left[ L{{T}^{-2}}\times L \right]}^{\dfrac{1}{2}}}$
Performing the multiplication and other functions in the equation will give,

& L{{T}^{-1}}={{\left[ {{L}^{2}}{{T}^{-2}} \right]}^{\dfrac{1}{2}}} \\\ & \Rightarrow L{{T}^{-1}}=L{{T}^{-1}} \\\ \end{aligned}$$ This means that the equation is dimensionally correct. Therefore the question has been answered. **Note:** It is mentioned that the object is falling freely under the gravitational force. Its velocity after traveling a particular distance $h$ is given as $V$. As the object is in free fall motion, the initial velocity will be zero. $u=0m{{s}^{-1}}$ The acceleration of the body will be given as, $a=g$ Velocity of the body after travelling a distance of $h$ will be given as $V$. Therefore the newton’s third law of motion is given by the equation, ${{V}^{2}}={{u}^{2}}-2as$ Where $s$ be the displacement of the body. Substituting the values mentioned in the question in this equation will give, ${{V}^{2}}={{0}^{2}}-2gh$ That is we can write that, $${{V}^{2}}=2gh$$ Taking the square root of this equation will give, $$V=\sqrt{2gh}$$ The square root of two can be taken separately, $$V=\sqrt{2}\times \sqrt{gh}$$ Therefore the value of $k$ will be, $$k=\sqrt{2}$$ This is the value of the constant.