Freshly picked cucumbers are dropped into a bin from a height of 1.25 m above the bottom of the bin. assuming that the bin is empty, how fast is a cucumber going when it hits the bottom of the bin? how much time does it take the cucumber to fall to the bottom of the bin?

a) Vf = 4.95 m/s

b) t = 0.51s

Explanation:

take downwards as positive.

let Vf be the final velocity as the cucumber reach the bottom of the bin and Vi be the initial velocity of the cucumber when they dropped.

a ) from equations of motion:

(Vf)^2 = (vi)^2 +2g×(x - x0) ,

since x0 = 0 m and Vi = 0 m/s.

Vf = \sqrt{2g×(x)}  = \sqrt{2(9.8)×(1.25)} = 4.95 m/s, downwards

therefore, the cucumber will reach the bottom of the bin with a speed of 4.95 m/s.

b) from equations of motion:

Vf = Vi + g×t

Vi = 0 then:

t = Vf/g = 4.95/9.8 = 0.51 s

therefore, the cucumber will take 0.51 seconds to reach the bottom of the bin.

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb