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.

An elevator whose floor-to-ceiling destance is 2.50 m starts ascending

1.	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» Solution :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`.