A ball is dropped from the top of a tower of 100 m height. Simultaneously another ball was thrown upward from the bottom of the tower with a speed of 50 m/s. They will cross each other after :  (g=10m/s²).

Height of the tower h = 100 m

Acceleration due to gravity g = 10 m/s²

Let the ball dropped from the top of the tower be P and the ball thrown upwards from the bottom of the tower be Q .

The initial velocity of ball P, u₁= 0 m/s

The initial velocity of ball Q, u₂= 50 m/s, 

Let the height from the top of the tower to the point of intersection be ${ℎ}_1$.

Let the height from the bottom of the tower to the point of intersection be ${ℎ}_2$.

Let the time taken for the intersection of the two balls be t.

Height ${ℎ}_1$: 

$\mathrm{s}= \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

$\Rightarrow {\mathrm{ℎ}}_1=0+\frac{1}{2}\times 10\times {\mathrm{t}}^2$

$\Rightarrow {\mathrm{ℎ}}_1=5{\mathrm{t}}^2$       ...eq. $\left(1\right)$

Height  ${ℎ}_2$ :

Since the motion of ball Q is against the gravity according to Newton's second equation of motion we get,

$\mathrm{s}= \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

$\Rightarrow {\mathrm{ℎ}}_2=50\mathrm{t}-\frac{1}{2}\times 10\times {\mathrm{t}}^2$

$\Rightarrow {\mathrm{ℎ}}_2=50\mathrm{t}-5{\mathrm{t}}^2$    ...eq.$\left(2\right)$

Both the balls will cover h₁ and h₂ distance before meeting.

Hence, upon adding equation $\left(1\right)$ and $\left(2\right)$ we get, 

${\mathrm{ℎ}}_1+{\mathrm{ℎ}}_2=5{\mathrm{t}}^2+25\mathrm{t}-5{\mathrm{t}}^2$

$\Rightarrow {\mathrm{ℎ}}_1+{\mathrm{ℎ}}_2=25\mathrm{t}$

$\Rightarrow 100=25\mathrm{t}$

$\Rightarrow \mathrm{t}=\frac{100}{25}=4 \mathrm{s}$