From the top of a tower of height 100m a 10gm block is dropped freely and a 6gm bullet is fired vertically upwards from the foot of the tower with velocity 100ms^–1 simultaneously. They collide and stick together. The common velocity after collision is (g = 10ms^–2)

Their relative acceleration is zero, and their relative acceleration is $v_r=$100 m/s. They consequently collide t seconds later.

$\mathrm{s}={\mathrm{v}}_{\mathrm{r}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{r}}{\mathrm{t}}^2$

$$\begin{gathered} 100=100\mathrm{t}+0 \\ \Rightarrow \mathrm{t}=1\sec \end{gathered}$$

After one second,

${\mathrm{V}}_{\mathrm{block}}=\mathrm{gt}=10$

${\mathrm{V}}_{\mathrm{bullet}}=100-\mathrm{gt}=90$

By conservation of momentum,

${\mathrm{m}}_{\mathrm{block}}{\mathrm{V}}_{\mathrm{block}}+{\mathrm{m}}_{\mathrm{bullet}}{\mathrm{V}}_{\mathrm{bullet}}={\mathrm{m}}_{\mathrm{combined}}{\mathrm{V}}_{\mathrm{combined}}$

$$\begin{gathered} 0.01\times 10-0.006\times 90=0.016\times {\mathrm{V}}_{\mathrm{combined}} \\ \Rightarrow {\mathrm{V}}_{\mathrm{combined}}=-27.5 \end{gathered}$$

Therefore, the common velocity after collision will be 27.5 m/s or higher.

Hence the correct answer is 27.5m/s.