Question

A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of

• A. $3s$
• B. $5s$
• C. $7s$
• D. $9s$

Answer 1

The correct option is B

$5s$

Step 1: Given data and assumptions:

Initial velocity, $u=0m{s}^{-1}$

Distance covered in first three seconds $=$Distance covered in last second

Let final velocity be $v$ and $t$ is the time taken of the fallen body.

Step 2: Formula used:

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

Step 3: Find the time of the fallen body under gravity.

Velocity at $\left(t-1\right)$ second, $v=g\times \left(t-1\right)$ $\left[g=accelerationduetogravity\right]$ ….$\left(i\right)$

Distance covered in last second.

$$\begin{gathered} \Rightarrow s=g\times \left(t-1\right)\times 1+\frac{1}{2}g\times 1^2\left[fromequation\left(i\right)\right] \\ \Rightarrow s=g\left(\frac{2t-1}{2}\right)......\left(ii\right) \end{gathered}$$

Distance covered in first three second.

$$\begin{gathered} \Rightarrow s=0\times 3+\frac{1}{2}\times g\times 3^2\left[initiallyvelocity,u=v=0\right] \\ \Rightarrow s=\frac{9}{2}g.......\left(iii\right) \end{gathered}$$

According to the question:-

Distance covered in first three seconds $=$Distance covered in last second

$$\begin{gathered} \Rightarrow g\left(\frac{2t-1}{2}\right)=\frac{9}{2}g\left[fromequation\left(ii\right)and\left(iii\right)\right] \\ \Rightarrow 2t-1=9 \\ \Rightarrow t=5s \end{gathered}$$

Hence, Option B is correct.