An elevator whose floor-to-ceiling destance is `2.50 m` starts ascending with a constant acceleration of `1.25 ms^(-2)` On second after the start, a bolt begins falling from the elevator. Calculate:
a. The free fall time of the bolt
b. The displacement and reference frame of ground.

Answer

`(2)/(3)`;
`(25)/(18)m; (445)/(288) m`

Explanation

a. ` t=sqrt((2h)/(g+a))=sqrt((2xx2.5)/(10+1.25))=(2)/(3) s `
b. Velocity of lift after `1` s.
`v=0 +1.25 xx 1=1.25 ms^(-1) =5//4 ms^(-1)`
Ths will be the initial velocity of bolt.
Destance moved up by lift in `2//3 s`,
` x=(5)/(4)xx(20/(3)+(1)/(2)1.25((2)/(3))^(2)=(10)/(9)m`
Displacement of bolt `=2.5-x=2.5 -(10)/(9) =(25)/(18) m`
Maximum height attained by bolt above the point of dropping ,
`(v^(2))/(2g)=((5//4)^(2))/(2xx 10) =(5)/(64) m`
So distance travelled=`2xx (5)/(64)+ (25)/(18)=(445)/(288)m`.