Asked by karan
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]
Answers
Answered by
bobpursley
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.
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.
Answered by
Ritesh upadhyay
40
Answered by
Ritesh upadhyay
48
Answered by
Mayukh
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
Answered by
Ayush
129
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