A lift moves upward with an acceleration of 1.2 ms -2 . A nail falls from the ceiling of the lift 3 m above the floor of the lift, Distance of its fall with reference to the shaft of the lift is
Text Solution
Verified by ExpertsC
Let y 1 be the distance covered by the fan and y 2 be the distance covered by lift just before the fan reaches the floor of the lift.
∴ ∴ y 1 + y 2 = 3
but y 1 = 2.4t + 1/2 9.8 t 2
(initial velocity of fan is upward and acceleration is downward)
and y 2 = 2.4t + 1/2 × 1.2 t 2
= 2.4t + 0.6t 2
∴ ∴ y 1 + y 2 = 4.9t 2 + 0.6t 2 = 5.5t 2
or 3 = 5.5 t 2
or t = 
= 0.74 s
Now y 1 = -2.4 × 0.74 + 4.9 × 0.74 2
= -1.68 + 2.68 = 1 m
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems