A ball of mass 100gm is projected vertically upward from the ground with a velocity of$50{\text{ms}}^{-1}$ . At the same time another identical ball is dropped from a height of 100m to fall freely along the same path as that followed by the first ball. After some time the two balls collide, stick together and finally fall to the ground. The time taken by the combined mass to fall to the ground is approximately $\left(g=10{\text{ms}}^{-2}\right)$

$\to \text{A}\to \text{B}$

$\text{S}=\text{ut}+\frac{1}{2}{\text{gt}}^{\text{2}}$                                   

$100-x=\frac{1}{2}{\text{gt}}^{\text{2}}\to \left(1\right)$

$\to \text{B}\to \text{C}$

$\text{S}=\text{ut}+\frac{1}{2}{\text{gt}}^{\text{2}}$

$x=50\text{t}-\frac{1}{2}{\text{gt}}^{\text{2}}\to \left(2\right)$

From (1) and (2)

$100=50\text{t}$

$\text{t}=2\sec$

$\to \text{v}=\text{u}+\text{at}$

${\text{u}}_1=\text{u}-\text{gt}$

$=\text{50}-\text{10}\left(2\right)$                                    

$=30\text{m/s}$

${\text{u}}_2=0+10\left(2\right)$                                

$=2\text{0m/s}$

$\text{v}=\frac{{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}+{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{{\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}}$

$=\frac{\text{m}\left(-30\right)+\text{m}\left(20\right)}{{\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}}$

$=\frac{-10\text{m}}{2\text{m}}=-5\text{m/s}$

$\Rightarrow {\text{h}}_{\max }=\frac{{\text{u}}^{\text{2}}}{\text{2g}}=\frac{{\left(-5\right)}^2}{2\left(10\right)}=1.25\text{m}\text{from}\left(2\right)$

$\text{h}=50\left(2\right)-\frac{1}{2}\left(10\right)\left(4\right)=80\text{m}$

Total height $=80+1.25$

$=81.25\text{m}$

$\to {\text{t}}_1=\sqrt{\frac{{\text{2h}}_{\text{0}}}{\text{g}}}$

$=\sqrt{\frac{2\times 81.25}{10}}$

${\text{t}}_1=4\sec$

${\text{t}}_{\text{2}}=\frac{\text{u}}{\text{g}}$

$=\frac{5}{10}$

${\text{t}}_2=0.5\sec$

Total time $=4+0.5$

$=4.5\sec$