# Calculate the time of flight of the bolt from ceiling to floor

Motion question - plz help

## Homework Statement

A lift ascends with an upward acceleration of 1.5ms^(-2). At the instant its upward speed is 2.0ms^(-1), a loose bolt drops from the ceiling of the lift 3.0m from the floor. Calculate:
(a) the time of flight of the bolt from ceiling to floor

## Homework Equations

x(f) = x(i) + V(xi)t + (1/2)a(x)t^(2)

final position = initial position + (initial velocity x time) + (0.5 x acceleration x time^(2))

## The Attempt at a Solution

I attempted to solve this question by forming 2 equations and solving it, since when the bolt leaves the roof of the lift and by the time it reaches the floor of the lift, the lift would've moved up:

BALL: x(f) = 0.5 x (-9.8) x (t^(2))

LIFT: x(f) = 3 + 2t + (0.5 x 1.5x t^(2))

answer: 0.73sec

from the two above equations I could't get the right answer, so can someone plz tell me if I've formed the right equations? Thanks in advance....

## Answers and Replies

Astronuc
Staff Emeritus
Science Advisor

Look at the problem statment carefully.

Initially, until the bolt becomes free, it is traveling with the lift, and it has the same upward initial velocity as the lift at the moment it becomes free.

So rewrite BALL: x(f) = 0.5 x (-9.8) x (t^(2)) with the correct initial conditions

Also realize that the initial position of the bolt is 3 m above the inital position of the floor.

Look at the problem statment carefully.

Initially, until the bolt becomes free, it is traveling with the lift, and it has the same upward initial velocity as the lift at the moment it becomes free.

So rewrite BALL: x(f) = 0.5 x (-9.8) x (t^(2)) with the correct initial conditions

Also realize that the initial position of the bolt is 3 m above the inital position of the floor.

Thanks for your reply, I never thought of that... Thanks again. :)