An elevator descends into a mine shaft at the rate of 9m per min. If it starts from 21m above the how long will it take to reach -15m
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Answers
Answer:
Given that,
Speed of the lift = 10 m/s
Distance traveled by the lift in 4 seconds = Speed of the lift × 4
d=10×4
d=40 m/s
Now, Initial Speed of the Stone u=30 m/s
Final Velocity of the stone v=0
Acceleration a=−g
The Stone is thrown against the gravitational force
So,
a=−10 m/s²
Using the third Equation of Motion,
v²−u²=2as
0−(30)²=2(−10)s
20×s=900
s=45m
Thus, Distance covered by the Stone in the certain time is 45 m.
∵ v−u=at
∴ 0−30=−10t
10t=30
t=3 sec
Thus, time taken by the Stone to reach the height of 45 m is 3 sec.
∴ Distance traveled by the lift in 3 seconds d=3×10=30 m
After Reaching the Height of 45 m, stone starts falling from there and covered the distance of (45+x) m
During the Fall,
Initial velocity u=0
Accelerationa=−10 m/s
2
Time taken by the stone to covers the distance (45+x) m during the fall = Time taken by the stone to travels the distance of x m.
Using the Second Equation of the motion,
s=ut+1/2at²
After solving the Equating,
The value of x will be 129 m.
Now,
3
x
=
3
129
3
x
=43 m
Hence, The value of
3
x
=43 m.
Step-by-step explanation:
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