Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms^$-$1. [use g = 10 ms^$-$2] :

As the meeting point lies $100 \mathrm{~m}$ above ground, displacement of ball will be $80 \mathrm{~m}$. <br/><br/>For ball $A$ <br/><br/>$$ \begin{aligned} &u=0, \mathrm{~S}=80 \mathrm{~m}, a =+\mathrm{g}=+10 \mathrm{~m} / \mathrm{s}^2 \text {, time }=t_1 \\\\ &\Rightarrow S =u t+\frac{1}{2} a t^2 \\\\ &\Rightarrow 80 =0+\frac{1}{2} \times 10 \times t_1{ }^2 \\\\ &\Rightarrow \frac{160}{10} =t_1{ }^2 \\\\ &\Rightarrow \quad t_1 =4 \mathrm{~s} \end{aligned} $$ <br/><br/>As ball $B$ is thrown after 2 seconds after release of $A$. Thus, time available for ball $B$ is 2 seconds to cover a distance of $80 \mathrm{~m}$. <br/><br/>Let speed be ' $u$ ' $\mathrm{m} / \mathrm{s}, t_2=4-2=2 \mathrm{~s}, \mathrm{~S}=80 \mathrm{~m}$, $a=+g=+10 \mathrm{~m} / \mathrm{s}^2$ <br/><br/>$\therefore 80 =u \times 2+\frac{1}{2} \times 10 \times(2)^2$ <br/><br/>$\Rightarrow 80 =u+20$ <br/><br/>$\Rightarrow 2 u =60$ <br/><br/>$\Rightarrow u =30 \mathrm{~m} / \mathrm{s}$