Ball A is dropped from rest from a building of height H exactly as ball B is thrown up vertically from
the ground. After some time ball A collide with ball B when ball A is moving upward direction. When A collide with B, ball A has twice the speed of B. This collision occurs at height h.
(i) The value of 
is–
Text Solution
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Ans.
(i)
Sol Y A = H – 
gt 2 Y B = V 0 t – 
gt
2 Y A = Y B = h
H = V 0 t C (t C = time at which collide)
= – 2g (Y A – H)
= V 0
2 – 2g Y B
= 4V B 2 V A = – gt C
V B = V 0 – gt C
(ii)
Sol Y A = H – 
gt
2 Y B = V 0 t – 
gt
2 Y A = Y B = h
H = V 0 t C (t C = time at which collide)
= – 2g (Y A – H)
= V 0
2 – 2g Y B
= 4V B 2 V A = – gt C
V B = V 0 – gt C
(iii)
Sol Y A = H – 
gt
2 Y B = V 0 t – 
gt
2 Y A = Y B = h
H = V 0 t C (t C = time at which collide)
= – 2g (Y A – H)
= V 0
2 – 2g Y B
= 4V B 2 V A = – gt C
V B = V 0 – gt C
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