Question

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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Answer 1

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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