particle is dropped from the top of the tower. During its motion it covers 9/25 th part of the tower in the last one sec. Find the height of the tower.
SOLVE IT USING NTH SECOND FORMULA, I'LL MARK YOU THE BRAINLIEST OTHERWISE YOU WON'T BE PROVIDED POINTS
SOLVE IT USING NTH SECOND FORMULA, I'LL MARK YOU THE BRAINLIEST OTHERWISE YOU WON'T BE PROVIDED POINTS
Answers & Comments
The correct option is C
123
m
Lets consider,
T
=
total time taken to reach the ground
H
=
height of the tower
Initial velocity of the particle is zero.
g
=
9.8
m
/
s
2
According to the question,
From the 2nd equation of motion,
9
H
25
=
1
2
g
(
T
−
1
)
2
. . . . . . .(1)
16
H
25
=
1
2
g
T
2
. . . . . . . . .(2)
Divide above two equation, we get
16
9
=
[
T
T
−
1
]
2
4
3
=
T
T
−
1
3
T
=
4
T
−
4
T
=
4
s
e
c
Substitute the value of
T
in equation (2),
16
H
25
=
1
2
×
9.8
×
4
×
4
H
=
25
16
×
78.4
=
123
m
The correct option is D.