. A ball is dropped from a bridge of 122.5 metre
above a river. After the ball has been falling for
two seconds, a second ball is thrown straight down
after it. Initial velocity of second ball so that both
hit the water at the same time is
(1) 49 m/s
(2) 55.5 m/s
(3) 26.1 m/s
(4) 9.8 m/s
Answers
Answer:
Hence,
option (C) is correct answer.
is right answer your questions
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
Hence,
option (C) is correct answer.
Answer:
option 3
Explanation;
Sa=Sb
GIVEN:
Sa=122.5m
Ua=0
ta=?
FROM EQUATION OF MOTION
S=UT+1/2gt^2
122.5=0+1/2*9.8*t^2
HENCE.....ta=5
given that after 2 sec a second ball is thrown,So
Tb=5-2=3
122.5=UbTb+1/2gTb^2
122.5=Ub*3+1/2*9.8*3^2
122.5=3Ub+1/2*9.8*9
122.5=3Ub+44.1
3Ub=122.5-44.1
3Ub=78.4
Ub=78.4/3
Ub=26.1
HENCE THE INITIAL VELOCITY OF THE SECOND BALL IS 26.1