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

The correct option is A $40\text{m}$
Given:
Wooden piece of mass $=0.03\text{kg}$
Building height $=100\text{m}$
Bullet mass $=0.02\text{kg}$
Initial velocity of bullet $=100{\text{ms}}^{-1}$,


Time taken for the particles to collide,

For wooden ball

Let $h_1$ be the distance which is travelled, then using equation of motion

$h_1=0+\frac{1}{2}g{t}^2$

Similarly, $h_2$ is the distance covered by the bullet in verticxal direction

$h_2=100.t-\frac{1}{2}g{t}^2$

Since, total distance covered by both,

$h_1+h_2=100\text{m}$

$100t=100\text{m}\Rightarrow t=1\text{s}$

Speed of wood just before collision

$u_1=0+g.t=10\text{m/s}$

and speed of bullet just before collision

$u_2=v-gt$$=100-10\times 1=90\text{m/s}$

Distance travelled by the bullet (collision point)
$S=100\times 1-\frac{1}{2}\times 10\times 1=95\text{m}$

Now, using conservation of linear momentum just before and after the collision
$m_1{u}_1+m_2{U}_2=(m_1+m_2)v$

$-\left(0.03\right)\left(10\right)+\left(0.02\right)\left(90\right)=\left(0.05\right)v$

$\Rightarrow 150=5v$

$\therefore v=30\text{m/s}$

Max. height reached by body,

$v^2=u^2-2gh$

$\Rightarrow h=\frac{v^2}{2g}=\frac{30\times 30}{2\times 10}=45\text{m}$

$\therefore$ Height above tower $45-5=40\text{m}$

Hence, option $\left(C\right)$ is correct.