The ball is dropped from top of a tower of 100 m height. Simultaneously another ball was thrown upward from the bottom of the tower with speed of $50{\text{ms}}^{-1}$ . They will cross each other after, [g =$10{\text{ms}}^{-1}$ ]              

Solution:

Given that

(i) A ball is dropped from a tower $\text{H}=100\text{m}$  

$\text{u}\text{=}\text{0}$

$a\text{=}\text{+}\text{g}\text{=}{\text{10m/s}}^{\text{2}}$

(ii) Another ball is projected up vertically at same time with$\text{u}=50{\text{ms}}^{\text{-1}}$  

$a=-\text{g}=-10{\text{m/s}}^{\text{2}}$  

Let these two balls cross each other at time t second

Then we get the height descended by ${\text{I}}^{\text{st}}$ ball

H the height ascended by ${\text{II}}^{\text{nd}}$  ball = height of tower

$\Rightarrow$ $\frac{1}{2}{\text{gt}}^{\text{2}}+\text{ut}-\frac{1}{2}{\text{gt}}^{\text{2}}=\text{H}$

$\therefore 50\text{t}=100$

$\therefore \text{t}=\frac{100}{50}=2\text{sec}$  

So they cross each other after$\text{t}=2\text{s}$                             

Then${\text{I}}^{\text{st}}$  ball descended through$=\frac{1}{2}\times {\text{gt}}^{\text{2}}=\frac{1}{2}\times 10\times 4$  $=20\text{m}$  

Second ball have ascended up second$=80\text{m}$