Question

a lift is descending with uniform acceleration . To measure the acceleration ,a person in the lift drops a coin at the moment when lift was descending with speed
6ft/sec .The coin is 5 ft above the floor of the lift at the time it is dropped the person observes that the coin strikes the floor in 1 sec . Calculate the acceleration of the lift in ft/s^2

Answer 1

According to the Question,

A lift is descending with a uniform acceleration let say it as a ft/s².

Now, When the coin is dropped by the Person in the lift, it falls freely under gravity.

∴ The acceleration of the coin with Respect to the Ground is same as the acceleration due to gravity, g (= 9.8 m/s²).

∴ Acceleration due to gravity = 9.8 m/s² = 32 ft/s².

∵ Acceleration of the the Lift is equals to the 
= Acceleration of Coin w.r.t the Ground - Acceleration of the Coin w.r.t. the lift.

∴ Acceleration of the Coin w.r.t. the lift = Acceleration of the Coin w.r.t. the Ground - Acceleration of the Lift.

∴ Acceleration of the Coin w.r.t. the lift = (g - a) ft/s².   
   = (32 - a)ft/s²   

From the Question, 

Time taken by the Coin to strikes the Ground (t) = 1 seconds.
Distance covered through the coin before striking the ground (S) = 5 ft.

 Initial Velocity(u) of the coin w.r.t. the lift = 0
[Since, the coin starts from the rest.]


From the Second Equation of the Motion,

S = ut + (1/2)a₁t².

Where,
a₁ = acceleration of the Coin w.r.t. the lift.

5 = 0 + (1/2)a₁(1)² 
5 = 0 + a₁/2 
a₁ =  10 
a₁ = 10  ft/s².

Now, Acceleration of the Coin w.r.t. the lift = 10 ft/s².

∴  10 = 32 - a 
  a = 32 - 10 
  a =  22 ft/s².

∴ Acceleration of the lift is 22 ft/s².


Hope it helps.