let hi be top of mine shaft
position of lift= -10t
position of stone=30t-4.9(t-4)^2
set these equal, solve for t.
then, solve for lift final postion at t.
A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s. 4 s later a boy throws a stone vertically upwards from the top of the shaft with a speed of 30 m/s. Find when and where stone hits the lift [take: g=10 m/s²]
5 answers
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For the stone to hit the lift, displacement of both the lift and the stone must be equal.
Let's say it hits in t second
30t-0.5gt²=-10t
Solve for t and it's the answer
Let's say it hits in t second
30t-0.5gt²=-10t
Solve for t and it's the answer
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