Ball A has dropped from a tower which is 200 m above ground level from rest ,at the same time a second ball is thrown from the ground upward at an initial velocity of 80 [m ⁄ s]. Find the time and altitude of the bodies when they are passing each other, and find the minimum initial velocity upward of ball B in order to enable such a meeting.
We acknowledge that at the meeting point the altitude h and the travel time of both balls are the same.
For ball A: 

(1) 
For ball B: 

(2) 
Comparing both equations (1) and (2) we get the time to encounter t:
The meeting point altitude is from eq. (2):
Total falling time of ball A to the ground calculated by eq. (1a) is:
The minimum upward velocity is when the two balls will meet at the ground level (see figure 3), it means that the maximum height of ball B will be reached after half the full falling time: t = 6.4 / 2 = 3.2 [s]
v_{0B} = v_{t} − g t = 0 + 9.8 * 3.2 = 31.4 [m / s]
Another approach to calculate the time t until encounter (Figure 2) is to calculate the difference of the distance between ball A and ball B as a function of time by:
∆x = ( H − h_{A}) − h_{B}
At the time of encounter of both balls ∆x = 0 > t = 200 / 80 = 2.5 [s] we get the same answer as before.
