A ball is projected vertically upward with an initial velocity of $50{ \mathrm{m} \mathrm{s}}^{-1}$ at $t=0 \mathrm{s}$. At $t=2 \mathrm{s}$, another ball is projected vertically upward with same velocity. At $t=$____$\mathrm{s}$, second ball will meet the first ball $\left(\mathrm{g}=10{ \mathrm{m} \mathrm{s}}^{-2}\right)$.

Distance covered by ball $A$ is $h=50\left(2\right)-\frac{1}{2}\times 10\times 2^2$$=100-20=80 \mathrm{m}$

Now, at time $t=2 \mathrm{s}$, velocity of $A$ is $v_A=50-10\times 2=30 \mathrm{m} {\mathrm{s}}^{-1}$

Relative velocity of both balls after $2 \mathrm{s}$ is $v_{\mathrm{rel}}=50-30=20 \mathrm{m} {\mathrm{s}}^{-1}$

Here, relative displacement $x_{\mathrm{rel}}=80 \mathrm{m} {\mathrm{s}}^{-1}$ and relative acceleration$a_{\mathrm{rel}}=0$.

Time taken $t=\frac{80}{20}=4 \mathrm{s}$

Time at which they will meet is $t=4+2=6 \mathrm{s}$