A lift is moving upwards with an acceleration of 2m/s^2. When the upward velocity of the lift is 5m/s, a particle is projected vertically upwards with a velocity of 3m/s with respect to the floor of the lift by a passenger. Find the time after which it reaches the passenger again
a) (2/3)^(1/2) sec
b)0.5 sec
c) (3/4)^(1/2) sec
d) 1 sec

Dear Student,

Please find below the solution to the asked query:

The relative velocity between the lift and the lift and the particle is $5+3-5=3$ m/s. Lift will move with acceleration $2 {\text{ms}}^{-2}$ upwards while due to gravity the particle will have $g=10 {\text{ms}}^{-2}$ downwards. Now the relative acceleration of the particle about the lift is $-10-2=-12 {\text{ms}}^{-2}$. Suppose the particle will come back after time t, then we have
$$\begin{gathered} ut-\frac{1}{2}a{t}^2=0 \\ \Rightarrow 3t-\frac{1}{2}\times 12\times t^2=0 \\ \Rightarrow t=\frac{6}{12}=0.5 \text{s} \end{gathered}$$
So option b) is the correct answer. 

Hope this information will clear your doubts about motion in a straight line.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards