Question

$solve \: this \: question$
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Answer 1

See the first figure

Actually, Ball A:

$total \: time \: travel \: t (1) \\ = \sqrt{ \frac{2h}{g} } = \sqrt{ \frac{2 \times 200}{10 } } = 2 \sqrt{10}s$

Ball B:

$total \: time \: of \: flight \: t(2) \\ = \frac{2u}{g} = \frac{2 \times 10}{10 } = 2s$

See the second figure

so they don't meet in first 2s.

See the third figure

After 2√10s , Ball-A reaches ground and

Ball-B remains at rest lying on the ground.

So, then they meet.

Therefore, Required time = 2√10s