Q: A ball is falling from the top of a cliff of height h with an initial speed v . Another ball is simultaneously projected vertically up with the same speed. When do they meet ?

[ Ans: h/2v ]

Sol: For downward motion

$ \displaystyle h_1 = v t + \frac{1}{2}g t^2$ …(i)

For upward motion

$ \displaystyle h_2 = v t – \frac{1}{2}g t^2$ …(ii)

Adding (i) & (ii)

$ \displaystyle h_1 + h_2 = v t + vt $

$ \displaystyle h = 2 v t $

$ \displaystyle t = \frac{h}{2 v} $