Question

a lift is moving up with a constant retardation of 2 ms2 when its upward velocity is 5 ms a boy in the lift tosses a coin imparting it an upward velocity of 3 ms wrt himself . his fingers at the moment of toss are midway between the floor and ceiling whose total height is 2 m . find displacement of coin when it hits the ground.

Answer 1

Velocity of the floor of the lift = v1 (t) = 5 - 2 t    m/s
v1(t) = 0 at t = 2.5 s.  Then the lift starts to come down or halts after 2.5 s.
Position of the floor at t = 0, is the reference for y ie., y = 0.

Displacement of the floor = y(t) = 5 t - t²   m
Displacement of the ceiling of the lift y2(t) = 2+5 t - t²   m

Velocity of the coin = v3(t) = 5+3 - g t = 8 - 10 t   m/s
Position of the coin: y3(t) = 1 + 8 t - 5 t²    m/s,  as y2(0) = 1 m

  y1(t) = y3(t)  ==> 5 t - t² =  1 + 8 t - 5 t²
                             4 t² - 3 t - 1 = 0
                            (4t + 1) (t - 1) = 0  => t = 1 sec.

y1(t=1) = 5*1-1² = 4 m.      So the coin flies 3 m when it hits the lift floor.

We can check if the coin ever touches the ceiling when tossed. In case the coin hits the ceiling, then the answer could be different.
       y2(t) = y3(t) =>   2+ 5 t - t² = 1 + 8 t - 5 t²
                                 4 t² - 3 t + 1 = 0
                                 as 3² - 4*4*1 < 0, there are no roots.

Hence the coin does not hit the ceiling and the answer is  3 m.