A body dropped from top of a tower fall through 60 m during the last two second of its fall. The height of tower is
(8 = 10
10 ms-2) (Single Choice Question)
a 95 m
b
60 m
с
80 m
d 90 m
=
Answers
Answered by
4
Answer:
In the case of free fall displacement of the motion is
S=
2
1
gt
2
let assume
nth second = t
n
(n-2)th second = t
n−2
given that (t
n
−t
n−2
)=2...(1)
Displacement in last two seconds,
S
(n,n−1)
th
=
2
1
g(t
n
−t
n−2
)
2
putting values in this equation
⟹40=
2
1
×10×(t
n
−t
n−2
)
2
⟹(t
n
−t
n−2
)(t
n
+t
n−2
)=8
⟹(t
n
+t
n−2
)=4...(2)
by solving eq 1 and eq 2 we get the values of
⟹t
n
=3
⟹t
n−2
=1
hight of the building
S=
2
1
gt
2
⟹S=
2
1
×10×3
2
⟹S=45m
Answered by
2
Answer:
option C is the correct answer for your question
Explanation:
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