Question

A particle is dropped from rest from a large height. Assume g to be constant throughout, the motion. The time taken by it to fall through successive distances of 1m each will be.

1. All equal, being equal to$\sqrt {2/g}$second.
2. In the ratio of the square roots of the integers 1, 2.3,
3. In the ratio of the difference in the square roots of the integers, i.e. $\sqrt 1 ,(\sqrt 2 - \sqrt 1 ),(\sqrt 3 - \sqrt 2 ),(\sqrt 4 - \sqrt 3 ),.....$
4. In the ratio of the reciprocals of the square roots of the integers, i.e. $\dfrac{1}{{\sqrt 1 }},\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt 3 }},....$

Answer 1

Here we have to apply the formula for kinematics. We have to take out the time taken; we only have been given with the constant gravity g and distance of 1m. Apply the formula $s = ut + \dfrac{1}{2}a{t^2}$ here's = Distance, u = Initial velocity, a = Acceleration, t = time taken. Assume for n meter then n+1, subtract the consecutive terms will we get the time taken.

Complete step-by-step answer:
Step 1:
Apply the formula for kinematics in which distance is included.
$s = ut + \dfrac{1}{2}a{t^2}$
Put the given value in the above equation and solve,
$n = 0 \times t + \dfrac{1}{2}g{t_n}^2$
Here the initial velocity of the particle is zero because it is dropped from a height.
$n = \dfrac{1}{2}g{t_n}^2$
Solve for${t_n}$,
$\sqrt {\dfrac{{2n}}{g}} = {t_n}$;
Now for ${t_{(n + 1)}}$
$\sqrt {\dfrac{{2(n + 1)}}{g}} = {t_{n + 1}}$;

Step 2: Calculate the time taken:
Subtract $\sqrt {\dfrac{{2(n + 1)}}{g}} = {t_{n + 1}}$from $\sqrt {\dfrac{{2n}}{g}} = {t_n}$
We have,
${t_{n + 1}} - {t_n} = \sqrt {\dfrac{{2(n + 1)}}{g}} - \sqrt {\dfrac{{2n}}{g}}$
Solve further,
${t_{n + 1}} - {t_n} = \dfrac{{\sqrt 2 }}{g}(\sqrt {n + 1} - \sqrt n )$
Now put the value of n from 0, 1, 2, 3, 4……m
We will get the value of time taken as a ratio:
$\sqrt 1 ,(\sqrt 2 - \sqrt 1 ),(\sqrt 3 - \sqrt 2 ),(\sqrt 4 - \sqrt 3 ),.....$.

Final Answer: The time taken by it to fall through the successive distances of 1m each will be $\sqrt 1 ,(\sqrt 2 - \sqrt 1 ),(\sqrt 3 - \sqrt 2 ),(\sqrt 4 - \sqrt 3 ),.....$

Note: Here don’t apply the distance-speed formula; it will not work, instead apply the equation of kinematic in which distance is included. Assume the distance n and then for (n+1) and find out the value of n in the terms we will get a ratio.