Question

a ball is dropped from the top of a building of height 80m . at same instant another ball is thrown upwards with speed 50m/s from the bottom of the building . the time at which balls will meet is ?

Answer 1

Suppose the ball will meet at t second.
The position of the first ball in t second will be $h=80-\frac{1}{2}g{t}^2$
and position of the second ball in t second will be $h=ut-\frac{1}{2}g{t}^2=50t-\frac{1}{2}g{t}^2$.
Equating this two we get
$$\begin{gathered} 80-\frac{1}{2}g{t}^2=50t-\frac{1}{2}g{t}^2 \\ t=\frac{80}{50}=1.6 \end{gathered}$$