An elevator is descending with uniform acceleration.To measure the acceleration, a person in the elevator drops a coin at momen the elevator strts. The coin is 6 ft asbove the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate these dta the acceleration of the elevator.

Explanation

Method 1: Observation from fround frame
Let us assume that the elevator has a velocity `v_(0)` when the coin loses contact with the elevator aat time `t. Let the coin stsrike the base of the elevator after time `t=T`, just below the point of loosing contact with froof of the coin.
In time `T`, the elevatior moves up the displacement `y` (upward) and the coin has net displacement `(h-y')=y` (downward)
.
For motion of coin:
Displacement of elevator during tim `=-y=-(h-y)`
Using equation `S=ut+(1)/(2) at^(2)`,
`-(h-y'=v_(0)T-(1)/(2) g T^(2)` (i)
For motion of elevator :
Displacement of elevator during time `T=+y'`
`=y'=v_(0)T+(1)/(2)aT^(2)`
Adding (i) and (ii), we have `h=(1)/(2)(g+a)T^(2)`
This yields `T=sqrt(2h)/(g+a)`
Method 2: Analyze the motion of coin with respect to the observer sstanding in the elevator. As the coin releases from rest inside elevator, its velocity with respect to ground id equal to the velocity of elevator.
Initial relative velocity of coin w.r.t observer in the elevator,
`[vec u_(coin)]_(elevator)=[vec u_(coin)]_(ground)-[vec u_(elevator)]_(ground)=u-u=0`
and acceleration of equation for relative motion
`[vec s_(coin)]_(elevator)=[vec u_(coin)]_(clevator)[vec u _(coin)]_(elevator)t+(1)/(2)[vec a_(coin)]_(elevator)t^(2)`
`-h=0 p-(1)/(2)(g+a)t^(2)`
This yilds `t=sqrt(2h)/(g+a)`.