Take upward as positive.

• Wood mass: $m_1 = 0.03\,\text{kg}$
• Bullet mass: $m_2 = 0.02\,\text{kg}$
• Building height: $100\,\text{m}$
• Bullet fired upward from ground with $u_2 = 100\,\text{m s}^{-1}$
• Wood is dropped from top, so $u_1 = 0$
• Acceleration for both: $a = -g = -10\,\text{m s}^{-2}$

Find the time when bullet and wood meet.

Position of wood from ground: $y_1 = 100 - \frac{1}{2}gt^2 = 100 - 5t^2$

Position of bullet from ground: $y_2 = 100t - \frac{1}{2}gt^2 = 100t - 5t^2$

At collision, $y_1 = y_2$: $100 - 5t^2 = 100t - 5t^2$ $100 = 100t$ $t = 1\,\text{s}$

Find velocities of wood and bullet at collision.

For wood: $v_1 = u_1 - gt = 0 - 10(1) = -10\,\text{m s}^{-1}$ (downward)

For bullet: $v_2 = u_2 - gt = 100 - 10(1) = 90\,\text{m s}^{-1}$ (upward)

Apply conservation of momentum during embedding.

Since bullet gets embedded in wood, collision is perfectly inelastic.

Common velocity just after collision = $V$.

$m_1 v_1 + m_2 v_2 = (m_1 + m_2)V$

$0.03(-10) + 0.02(90) = 0.05V$

$-0.3 + 1.8 = 0.05V$ $1.5 = 0.05V$ $V = 30\,\text{m s}^{-1}$

So the combined system moves upward with speed $30\,\text{m s}^{-1}$.

Find the height of collision.

Substitute $t=1$ s into either position equation: $y = 100 - 5(1)^2 = 95\,\text{m}$

So collision happens at $95\,\text{m}$ above ground, i.e. $5\,\text{m}$ below the top of the building.

Find how much further the combined system rises after collision.

At highest point, final velocity becomes zero: $0 = V^2 - 2gh$ $0 = (30)^2 - 2(10)h$ $900 = 20h$ $h = 45\,\text{m}$

Thus, after collision it rises $45\,\text{m}$ above the collision point.

Find height above the top of the building.

Since collision occurred $5\,\text{m}$ below the top, net rise above the top is: $45 - 5 = 40\,\text{m}$

Check options.

• A: $30\,\text{m}$
• B: $40\,\text{m}$
• C: $20\,\text{m}$
• D: $10\,\text{m}$

Hence the correct option is B.