For ball 1
\(S_1 = u_1t+ \frac12 a_1 t^2\)
\(200 - x = \frac12(10)t^2\) .......(1)
For ball 2
\(S _2 = u_2t+ \frac12 a_2 t^2\)
\(- x = -20t + \frac12 \times 10t^2\)
\(x = 20 t - \frac 1210t^2\) ......(2)
From equation (1) and (2)
\(200 - 20t+ 5t^2 = 5t^2\)
\(20t = 200\)
\(t = \frac{200}{20}\)
\(t = 10 sec\)
Putting the value of t in equation (1)
\(200- x = 5t^2\)
\(200-x = 5\times 100\)
\(200- 500 = x\)
\(x = - 300m\)
They will meet 300m above the ground.
