Question

A ball falls from height $h$. After $1s$, another ball falls freely from a point $20m$ below the point from where the first ball falls. Both of them reach the ground at the same time. The value of $h$ is

• A. $11.2m$
• B. $21.2m$
• C. $31.2m$
• D. $41.2m$

Answer 1

The correct option is C

$31.2m$

Step 1: Given Data

Height of the first ball ball $h_1=h$

Height of the second ball ball $h_2=h-20$

Let the time taken by the first ball be $t_1=t$.

Therefore, the time is taken by the second ball $t_2=t-1$

Let the acceleration due to gravity be $g=10m/s^2$.

Step 2: Formula Used

According to newton's second equation of motion

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

Since the ball starts from rest the equation can be rewritten for the first case as,

$h=\frac{1}{2}g{t}^2$ $\left(1\right)$

Similarly, for the second case

$h-20=\frac{1}{2}g{\left(t-1\right)}^2$

$\Rightarrow h=20+\frac{1}{2}g{\left(t-1\right)}^2$

Step 3: Calculate time taken

Upon substituting this in equation $\left(1\right)$ we get,

$\therefore \frac{1}{2}g{t}^2=20+\frac{1}{2}g{\left(t-1\right)}^2$

$\Rightarrow g{t}^2=40+g\left(t^2-2t+1\right)$

$\Rightarrow g{t}^2=40+g{t}^2-2gt+g$

$\Rightarrow 2gt=40+g$

$\Rightarrow t=\frac{40+g}{2g}$

$\Rightarrow t=\frac{40+10}{2\times 10}$

$\Rightarrow t=\frac{50}{2\times 10}=2.5s$

Step 4: Calculate the height

Upon substituting this in equation $\left(1\right)$ we get,

$h=\frac{1}{2}\times 10{\left(2.5\right)}^2$

$=31.25m$

Hence, the correct answer is option (C).