Question

If a particle is thrown vertically upwards, find the ratio of time taken to traverse the lower half to upper half of the journey to the highest point.
A. $1:1$
B. $2:1$
C. $\sqrt 2 :1$
D. $\sqrt 2 - 1:1$

Answer 1

Hint: The given question can be solved using the equations of motion. First, we need to find out the initial and final velocity. Then figure out the time taken to traverse the first half followed by the time taken to traverse the second half of the journey. Then the ratio of both has to be figured out.

Complete step by step answer:
Let us assume that the particle goes up to the height of h meter and the initial velocity to be $u$. The final velocity will be zero here, so using the equation of motion, we have –
${v^2} - {u^2} = 2as$
Now putting the value of v, final velocity and $a = - g$
${0^2} - {u^2} = 2( - g)h \\
\Rightarrow u = \sqrt {2gh} $

Now, at the half distance of the journey, the velocity will be-
Using the same equation of motion, i.e.
${v^2} - {u^2} = 2as$
And putting the values, $u = \sqrt {2gh} $, $a = - g$ and $s = h/2$
${v^2} - 2gh = 2( - g)\left( {\dfrac{h}{2}} \right) \\
\Rightarrow {v^2} = 2gh - gh = gh $
And calculating the value of v, i.e. $v = \sqrt {gh} $

Now, as we have the value of initial and final velocity, the time taken to traverse the first half of the journey will be –
Using the equation of motion, i.e. $v = u + at$
And putting the previously calculated value of $v = \sqrt {gh} $ , $u = \sqrt {2gh} $ and $a = - g$. Calculating the value of t, time for first half of journey,
$\sqrt {gh} = \sqrt {2gh} + ( - g)t \\
\Rightarrow t = \sqrt {\dfrac{h}{g}} (\sqrt 2 - 1) $

For the second half of the journey, the time is –
Using the equation of motion, i.e. $v = u + at$ for calculating the time for the second half of the journey, and putting values $v = 0,u = \sqrt {gh} ,a = - g$
$0 = \sqrt {gh} - gT \\
\Rightarrow T = \sqrt {\dfrac{h}{g}} $
Therefore, the ratio of the time taken for covering the first half of the distance to the time take in covering the second half of the distance is given by-
$\dfrac{t}{T} = \dfrac{{\sqrt {\dfrac{h}{g}} (\sqrt 2 - 1)}}{{\sqrt {\dfrac{h}{g}} }} \\
\therefore \dfrac{t}{T} = (\sqrt 2 - 1):1 $

Hence, the correct answer is D.

Note: During the vertical movement of the object, it experiences a downward pull that causes a change in its velocity. As the object moves upward, the velocity decreases subsequently and becomes zero at the topmost point. The decrease in velocity is inversely proportional to the height. The acceleration remains the same throughout the journey, i.e. -g, which is also known as the acceleration due to gravity.