An elevator car whose floor of ceiling distance is equal to 2.7 m starts ascending with constant acceleration of 1.2 m per second square 2 second after the start a built begins falling from the ceiling of the car the free fall time of the bolt is
Answers
Answered by
1
Answer:
Let the total time of free fall be t
Then at t=2s
4
B
=2.
m
1
/s 4
C
=2.
m
1
/s
a
C
=1.2 m/s
2
a
B
=−10 m/s
2
Wrt car.
distance travelled by bolt =27 m
Then
−2.7 m=(4
B/C
)t+a
B/C
(
2
t
2
)
⇒−2.7=(2.4−2.4)t(
2
−10−12
)t
2
⇒2.7=
2
11.2
t
2
⇒t
2
=(
11.2
2.7
×2)
−11
=0.482 s
2
Note: 4
B
=4
C
till t=2.3 before the both spots to fall.
4
B
=4
C
=a
C
t=(1.2)(2)=2.4m/s
let the displacement of the bolt be h Then,
h=4
B
(t)+4
2
B
t
2
h=(2.4)
0.482
2
−(10)
(0.482)
h=−0.743
h=≈−0.7 m
∣h∣≈0.7 m
Similar questions