An elevator moves downward in a tall building at a constant speed of 5.70 m/s. Exactly 4.95 s after the top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest. (a) At what time does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at t = 0 s.)(b) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume 1 floor congruent 3 m.)

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lcmendozaf lcmendozaf · Answer 1 · 2019-09-18T02:41:54+02:00

Answer:

a) t = 3.01s

b) 15th floor

Explanation:

First we need to know the distance the elevator has descended before the bolt fell.

[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]

Now we can calculate the time that passed before both elevator and bolt had the same position:

[tex]Y_{b}=Y_{e}[/tex]

[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]

[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex] Solving for t:

t1 = -1.87s t2 = 3.01s

In order to know how the amount of floors, we need the distance the bolt has fallen:

[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex] Since every floor is 3m:

Floors = Yb / 3 = 15 floors.