Question

The distance through which a body falls in nth second is h. The distance through which it falls in the next second is

• A. h
• B. h+g/2
• C. h-g
• D. h+g

Answer 1

The correct option is D

h+g

Step 1: Given data:

The distance through which a body falls in nth second is h

Step 2: From the equations of kinematics we have:

$s=ut+\frac{1}{2}a{t}^2$

Here,

s is the distance

u is the initial velocity (u = 0)

t is the time

a is the acceleration

Now, distance in nth second = $\frac{1}{2}g{n}^2(gistheaccelerationduetogravity)$

Distance in (n+1)th second = $\frac{1}{2}g{(n+1)}^2$

Step 3: Calculating the distance covered in the next second:

Let the distance through which it falls in nth second be y, then

$$\begin{gathered} \mathrm{y}=\mathrm{distance}\mathrm{in}(\mathrm{n}+1)\mathrm{th}\sec -\mathrm{distance}\mathrm{in}\mathrm{nth}\sec \\ y=\frac{1}{2}g{(n+1)}^2-\frac{1}{2}g{n}^2 \\ y=\frac{g}{2}(2n+1) \end{gathered}$$

Step 4: Also, from equations of kinematics, the distance in nth second is given by,

$$\begin{gathered} s_n=u+\frac{a}{2}(2n-1) \\ Inthiscase,u=0anda=g \\ {s}_n=\frac{g}{2}(2n-1) \\ h=\frac{g}{2}(2n-1) \\ Now,h+g=\frac{g}{2}(2n-1)+g \\ h+g=\frac{g}{2}(2n+1)=y \\ Therefore, \\ y=h+g \end{gathered}$$

Therefore, the distance covered in the next second is ‘(h + g)’

Hence, the correct option is (D).