Answer:
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) ,
<em>since x0 = 0 m and Vi = 0 m/s.</em>
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.