Question

$solve \: this \: question$
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Answer 1

See the first figure

See the first figureActually, 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

Answer 2

To find the height h reached by the ball thrown vertically, we need to use the formula "v2 = u2 - 2gh ",
with final velocity v = 0, initial velocity u = 10 m/s and g = 10 m/s2

hence h = u2 / 2g = (10×10) /(2×10) = 5 m

Time t to reach the highest point will be obtained from the formula " v= u-gt ", with v=0; hence time t = u/g = 10/10 = 1 s ;

After reaching the height h, the ball falls to the ground under gravity.

Time to reach ground will be obtained from the formula " S = ut+(1/2)gt2 ", here initial velocity u = 0 and S = (200+5) m; let this time be t1

hence 205 = (1/2)×10×t12 ; we get t1 = begin mathsize 12px style square root of 41 end style .

hence total time taken by the ball that is thrown vertically = ( begin mathsize 12px style square root of 41 end style +1 ) s ......................(1)
Let us now calculate the time taken by the ball which is thrown downwards from building height with speed 10 m/s .

To find the time, we need to use "S = ut+(1/2)gt2 " , with S = 200 m, and u = 10 m/s.

200 = 10×t + (1/2)×10×t2 or t2 + 2×t - 40 = 0 ..............(2)

solving the above equation (2) for t, we get t = ( begin mathsize 12px style square root of 41 end style -1 ) s .................(3)

If we find the difference between the time durations given in eqn.(2) and eqn.(3), we find the difference of 2 s

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Instead of doing the elaborate calculation as given above, we can get the answer in an alternate way.
when the ball thrown with speed 10 m/s from top of the building, it reaches the same point during its vertical fall with speed 10m/s.
Hence both the balls are at same speed at the building height during their vertical fall. Hence time duration to reach the ground will be same.
Only difference is, the ball which is thrown vertically travel 1 s to reach the maximum height and another 1 s to reach the top of building when it falls downwards.
Hence only this 1+1 = 2 s difference between the journey of two balls.