Question

A ball is dropped from the roof of a tower of height $h$. The total distance covered by it in the last second of its motion is equal to the distance covered by it in first $3s$. What is the value of h?

• A. $125m$
• B. $5m$
• C. $58m$
• D. $250m$

Answer 1

The correct option is A

$125m$

Step 1: Formula used:

$$\begin{gathered} v=u+at \\ {v}^2=u^2+2aS \\ S=ut+\frac{1}{2}a{t}^2 \end{gathered}$$

Here, we have

$$\begin{gathered} u=initialvelocity \\ v=finalvelocity \\ t=time \\ S=displacement \end{gathered}$$

Also the distance covered in $n^{th}$ second is given by an equation:

$S_n=u+\frac{g}{2}(2n-1)$

Step 2: Calculating the value of height $\text{'}h\text{'}$:

Using the third equation of motion, the distance covered by a ball in first $3s$ is given by,

$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow S=\frac{1}{2}\times 9.8\times {\left(3\right)}^2\because u=0 \\ \Rightarrow S=44.1m \end{gathered}$$

If ball takes $\text{'}n\text{'}$ seconds to reach the ground, then distance covered in $n^{th}$ second is given by,

$$\begin{gathered} S_n=u+\frac{g}{2}(2n-1) \\ \Rightarrow S_n=\frac{g}{2}(2n-1)\because u=0 \\ \Rightarrow S_n=\frac{9.8}{2}(2n-1) \\ \Rightarrow S_n=9.8n-4.9 \end{gathered}$$

Now according to the question, we have,

$$\begin{gathered} S_n=44.1 \\ \Rightarrow 9.8n-4.9=44.1 \\ \Rightarrow 9.8n=49 \\ \Rightarrow n=5s \end{gathered}$$

Therefore, height $\text{'}h\text{'}$is given by,

$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow h=\frac{1}{2}\times 10\times {\left(5\right)}^2\because S=h,andg=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \\ \Rightarrow h=125m \end{gathered}$$

Hence, option(A) is correct answer.