Question

an elevator descends into a mine shaft at the rate of 6 m per minute if the descend starts from 10 metre above the ground level how long will take to reach the shaft 350m below the ground level

Answer 1

ACC. to second eqn of motion,
S=ut + 1/2 at^2
[ TAKE DOWNWARDS AS -ve AND UPWARDS As +ve]
S= dsplacement= final - initial =-350 - (10) = -360 m
U= Initial velocity = 6 m / min =0.1 m/sec
a = -g = -10 m/sec2

Apply the above bold eqn..
-360 = 0.1 t -1/2 gt^2

SOLVE IT AND U WILL GET UR t

Answer 2

Answer:

ACC. to second eqn of motion,

S=ut + 1/2 at^2

[ TAKE DOWNWARDS AS -ve AND UPWARDS As +ve]

S= dsplacement= final - initial =-350 - (10) = -360 m

U= Initial velocity = 6 m / min =0.1 m/sec

a = -g = -10 m/sec2

Apply the above bold eqn..

-360 = 0.1 t -1/2 gt^2

SOLVE IT AND U WILL GET UR t

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Step-by-step explanation: