Question

A lift in a mine moves down 24m in 87mins. If it moves in uniform rate, find at what distance below the surface, it will be after 6mins. If the lift starts from a height 10m above the ground, find how deep the lift will go from the surface after 70mins?

Answer 1

Answer:

a) 1 655 m under ground b) 9.31 m under ground

Step-by-step explanation:

• V = 24/87 m/min
• V = 0.275 m/min
• T = 6 min

S = V × T

S = 0.275 × 6

S = 1.655 m

b) same velocity but time is 70 min and lift starts from a height of 10 m above ground

S = V × T

S = 0.275 × 70

S = 19.31 m

hence we know that the lift starts from a heigt of 10 m from the ground

so. S = 19.31 -10

S = 9.31 m