A ball is throuwn with a speed of 20 m//s from top of a buling 150 m high and simultaneously another ball is thrown vertically upward witn a speed of 30 m/s from the foot of the builing . Find the time when both the balls will meet `(g =10 m//s^2)`

Explanation

To solve the problem of when the two balls meet, we need to analyze their motions separately and then combine the equations. Let's break down the steps:

### Step 1: Define the variables
- Let \( h = 150 \, \text{m} \) (height of the building).
- Let \( u_a = 20 \, \text{m/s} \) (initial velocity of the ball thrown downwards).
- Let \( u_b = 30 \, \text{m/s} \) (initial velocity of the ball thrown upwards).
- Let \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
- Let \( t \) be the time in seconds when both balls meet.
...