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]

Answer 1

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.

Answer 2

40

Answer 3

48

Answer 4

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

Answer 5

129