A ball is dropped from a bridge 122.5 m above a river.After the ball has been falling for 2 sec, a second ball is thrown straight down after it .What must be the initial velocity of second ball so that both hit the water at the same time?
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Answer:
Explanation:
For the first ball: Let it takes
′
t
′
sec to river.
⇒122.5=
2
1
×g×t
2
------------------------(1)
for second ball: time taken =t−2sec.
122.5=u(t−2)+
2
1
g(t−2)
2
--------------(2)
⇒
2
gt
2
=ut−2u+
2
gt
2
+2g−2gt
⇒(2g−u)t=2g−2u
⇒t=
2g−u
2g−2u
from(1)
t
2
=
g
2×122.5
t=
98
2×122.5
=
7
35
=5sec
⇒10g−5u=2g−2u
⇒8g=3u
u=
3
8×9.8
=26.1m/s
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