a lift is moving downward with uniform acceleration to measure the acceleration a person in the lift drops a coin 6 ft above the floor of the lift . the coin strikes the floor in 1 second find the acceleration

The "effective" acceleration rate, a' , (measured positive downward) as observed in the elevator

satifies the equation
6 = (1/2)a' t^2 = a'/2
since t = 1 second to fall 6 feet)
Therefore a' = 12 ft/s^2

Without elevator acceleration, the observed downward acceleration rate would be g = 32.2 ft/s^2

Since a' = g -a,
where a is the true acceleration rate (downward),
a = 32.2 - 12 = 20.2 ft/s^2

To find the acceleration of the lift, we can use the equations of motion for an object moving with constant acceleration.

First, let's identify the given variables:
- Initial velocity (u) = 0 (since the coin was dropped)
- Final velocity (v) = ?
- Distance (s) = 6 ft
- Time (t) = 1 second
- Acceleration (a) = ? (what we need to find)

The equation we need to use is:

s = ut + (1/2)at^2

Plugging in the given values:

6 = 0(1) + (1/2)a(1^2)

Simplifying the equation gives us:

6 = (1/2)a

Multiplying both sides by 2:

12 = a

So, the acceleration of the lift is 12 ft/s^2 downward.