Question 167 of 308

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

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Answer: 6

Distance covered by ball $A$ is $h=50\left(2\right)-\dfrac{1}{2}\times 10\times {2}^{2}$$=100-20=80m$
Now, at time $t=2s$, velocity of $A$ is ${v}_{A}=50-10\times 2=30m{s}^{-1}$
Relative velocity of both balls after $2s$ is ${v}_{rel}=50-30=20m{s}^{-1}$
Here, relative displacement ${x}_{rel}=80m{s}^{-1}$ and relative acceleration${a}_{rel}=0$.
Time taken $t=\dfrac{80}{20}=4s$
Time at which they will meet is $t=4+2=6s$