Question

A body released from the top of a tower falls through half the height of the tower in $3s$. It will reach the ground after nearly

• A. $3.5s$
• B. $4.24s$
• C. $4.71s$
• D. $6s$

Answer 1

The correct option is B

$4.24s$

Step 1: Given data

1. The time taken in the first half is $3\mathrm{s}$
2. The initial speed of the body is, $\mathrm{u}=0{\mathrm{ms}}^{-1}$

Step 2: Calculating height

Let the height of the tower is h meter, and after t seconds the body will reach the ground.

Now, from the formulae,

$S=ut+\frac{1}{2}g{t}^2$ ………..(1)

Where “s” is the distance traveled by a body at a time “t”, “u” is the initial velocity and “g” is the acceleration due to gravity.

So,

$$\begin{gathered} h=\left(0\times t\right)+\frac{1}{2}\times \left(9.8\right)\times \left(3^2\right) \\ orh=44.1m \end{gathered}$$

Now, the height of the body is

$$\begin{gathered} H=2\times h=2\times 44.1 \\ orH=88.2m \end{gathered}$$

Step 3: Calculating time

Again, from the equation, $S=ut+\frac{1}{2}g{t}^2$

$$\begin{gathered} 88.2=\frac{1}{2}\times \left(9.8\right)\times t^2 \\ or{t}^2=\frac{88.2}{9.8}\times 2=18 \\ ort=4.24 \end{gathered}$$

So, after $4.24$ seconds the body will reach the ground. Therefore option (B) is correct