A piece of wood of mass $0.03kg$ is dropped from the top of a $100m$ height building. At the same time, a bullet of mass $0.02kg$ is fired vertically upward, with a velocity $100m{s}^{-1}$, 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 $(g=10m{s}^{-2})$,

The correct option is D $40m$
Draw diagram.

Find the time when particles will collide.

Formula used: $v=\frac{d}{t}$

Given,
Mass of wood $m=0.03kg$
Height of the building $H=100m$
Mass of the bullet $m=0.02kg$
Velocity of bullet $v=100m/sec$
Time taken for the particles to collide,

$t=\frac{d}{V_{rel}}$

$d=$ distance covered
$V_{rel}$ is the relative velocity

$=\frac{100}{100}=1sec$

Find the speed of bullet and wood.

Formula used: $v=u+at$
Speed of wood just before collision,

$=gt=\frac{10m}{s}$

Speed of bullet just before collision,

$v-gt=100-10=90m/s$

Find the velocity of combined system after embedded.

Formula used:

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

Conversation of linear momentum just before and after collision,
Total momentum before collision $=$ Total momentum after collision

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

$m_{\text{total}}{v}_c=m_b{v}_b+m_w{v}_w$

$m_{\text{total}}$ is combined mass of bullet and wood that is $0.05$
$v_c$ is the velocity combination,
$m_b$ is the mass of bullet,
$v_b$ is the velocity of bullet,
$m_w$ is the mass of wooden block
$v_w$ is the velocity of wooden block
Substitute the values, we get

$\Rightarrow \left(0.03\right)(-10)+\left(0.02\right)\left(90\right)=\left(0.05\right)v_c$
$\Rightarrow -\left(0.03\right)\left(10\right)+\left(0.02\right)\left(90\right)=\left(0.05\right)v_c$
$\Rightarrow 150=5v_c$
$\Rightarrow v_c=30m/s$

Find the maximum height.

Formula used: $h_{max}=\frac{v^2}{2g}$

Maximum height reached by body, Before collision:

$0.03kg\downarrow 10m/s$
$0.02kg\uparrow 90m/s$

After collision: $0.05kg\uparrow 30m/s$

$\Rightarrow h_{max}=\frac{30\times 30}{2\times 10}=45m$

height to which the combined system reaches above the top of the building$=45-5=40m$


Final answer: (c)