Question

A ball dropped from a point A falls down vertically to C, through the mid-point B. The descending time from A to B and that from A to C are in the ratio
A. $1:1$
B. $1:2$
C. $1:3$
D. $1:\sqrt 2 $

Answer 1

Hint:In this question we will assume some distance between A and B and apply Newton's equation of motion. Apply equation of motion for both cases i.e., From A to B and A to C, taking the ratios we will get the desired ratio.

Formula used:
Newton’s equation of motion,
$s = ut + \dfrac{1}{2}g{t^2}$
Here, $s$ is the displacement of the particle, $u$ is the initial velocity, $g$ is the acceleration of gravity and $t$ is the time period.

Complete step by step solution:
Let ${t_1}$ be the descending time for travelling from A to B and ${t_2}$ be the descending time for travelling from A to C. Assume the distance between A and B is $s$. In the question, we are given that a ball is dropped from a point A falls down vertically to C, through the mid-point B. It means the B is the midpoint of distance A and C. Descending time and height relation is given by Newton’s law of equation of motion.

Since the ball is dropped the initial velocity will be zero ($\therefore u = 0$).
So, for point A to B equation of motion can be written as
$ \Rightarrow s = \dfrac{1}{2}g{t_1}^2$ ---------(1)
Also, for point A to B equation of motion can be written as
$ \Rightarrow 2s = \dfrac{1}{2}g{t_2}^2$ ----------(2)
Now dividing equation 1 from 2 we get,
$ \Rightarrow \dfrac{s}{{2s}} = \dfrac{{\dfrac{1}{2}gt_1^2}}{{\dfrac{1}{2}gt_2^2}} \\ $
$ \therefore \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{1}{{\sqrt 2 }}$
Hence the ratio is $1:\sqrt 2 $ .

Hence, option D is correct.

Note: All we need to know in this case is Newton's equation of motion. Knowing this and making the straightforward substitutions required will result in the solution. Also keep in mind the distance between A and B and A and C. Do not get confused.