Question Number 16623 by Tinkutara last updated on 24/Jun/17
![A ball is thrown up in a lift with a velocity u relative to the lift. If it returns to the lift in time t, then acceleration of the lift is [Answer: ((2u − gt)/t) upwards]](Q16623.png)
$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{up}\:\mathrm{in}\:\mathrm{a}\:\mathrm{lift}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:{u}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}.\:\mathrm{If}\:\mathrm{it} \\ $$$$\mathrm{returns}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{in}\:\mathrm{time}\:{t},\:\mathrm{then} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\left[\boldsymbol{\mathrm{Answer}}:\right. \\ $$$$\left.\frac{\mathrm{2}{u}\:−\:{gt}}{{t}}\:\mathrm{upwards}\right] \\ $$
Answered by ajfour last updated on 24/Jun/17
$$\:\mathrm{s}_{\mathrm{lift}} =\mathrm{s}_{\mathrm{ball}} \:\mathrm{in}\:\mathrm{time}\:\mathrm{t}\:\:\:\:\left(\mathrm{upward}\:+\mathrm{ve}\right) \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{at}^{\mathrm{2}} \:=\:\mathrm{ut}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:\:\mathrm{a}\:=\:\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} }\left(\frac{\mathrm{2u}−\mathrm{gt}}{\mathrm{2}}\right)\mathrm{t}\:=\:\frac{\mathrm{2u}−\mathrm{gt}}{\mathrm{t}}\:\:\left(\uparrow\right)\:. \\ $$
Commented by Tinkutara last updated on 24/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$