A piece of wood of mass $0.03kg$ is dropped from the top of a building $100m$ high. At the same time, a bullet of mass $0.02kg$ is fired vertically upward with a velocity of $100m{s}^{-1}$ from the ground the bullet gets embedded in the wooded piece after striking. Then the maximum height to which the combined system reaches above the building before falling below is: ( take $g=10m{s}^{-2}$ ).

The correct option is A

$40$

Step 1. Given Data:

Mass of wood, $m_1=0.03kg$

Mass of bullet, $m_2=0.02kg$

The velocity of the bullet, $u=100m{s}^{-1}$

$g=10m{s}^{-2}$

Height of the building, $d=100m$

Step 2. Find time taken, speed of wood and speed of bullet:

Time taken by the particles to collide,

$\begin{array}{rcl}t& =& \frac{d}{V_b}\\ t& =& \frac{100}{100}\\ t& =& 1sec\end{array}$

Before the collision, the speed of wood

$\begin{array}{rcl}{v}_1& =& gt\\ & =& 10\times 1\\ & =& 10m{s}^{-1}\end{array}$

Before the collision, the speed of the bullet

$\begin{array}{rcl}{v}_2& =& u-gt\\ & =& 100-10\\ & =& 90m{s}^{-1}\end{array}$

Step 3. Conservation of linear momentum just before and after the collision:

$m_1{v}_1+m_2{v}_2=mv$

$\begin{array}{rcl}(0.03)\left(10\right)+(0.02)\left(90\right)& =& (0.05)v\\ 150& =& 5v\\ v& =& 30m{s}^{-1}\end{array}$

Step 4. Find distance covered by body,

$$\begin{gathered} h=\frac{v^2}{2g} \\ h=\frac{(30\times 30)}{2\times 10} \\ h=45m \\ S=ut+\left(\frac{1}{2}\right)\times 10 \\ =0+\left(\frac{1}{2}\right)10 \\ =5m \end{gathered}$$

Step 5. calculate the maximum height:

So, the height above the tower $$\begin{gathered} =45-5 \\ =40m \end{gathered}$$

Hence, the correct option is (A).