Two bodies are thrown upwards with same initial velocity 40 ms^-1  with time gap 4 sec. After what time from projection of first body they will meet  (g  = 10 m/s²)

$$\begin{gathered} \text{ Displacement covered by first object in }\mathrm{t} \mathrm{secs}, \\ {\mathrm{s}}_1=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^2=40\mathrm{t}-\frac{1}{2}10\times {\mathrm{t}}^2 \\ \text{ Displacement covered by second object in }(\mathrm{t}-4)\mathrm{secs}, \\ {\mathrm{s}}_2=40(\mathrm{t}-4)-\frac{1}{2}10\times (\mathrm{t}-4{)}^2 \\ \text{ Since they meet }{\mathrm{s}}_1={\mathrm{s}}_2 \\ \text{ We get }\mathrm{t}=6\text{ secs } \end{gathered}$$