Abolt is dropped from a bridge under construction, falling 97 m to the valley below the bridge. (a) how much time does it take to pass through the last 11 % of its fall? what is its speed (b) when it begins that last 11 % of its fall and (c) just before it reaches the ground?

The first thing we have to do for this case is write the kinematic equationsto
 vf = a * t + vo
 rf = a * (t ^ 2/2) + vo * t + ro
 Then, for the bolt we have:
 100% of your fall:
 97 = g * (t ^ 2/2)
 clearing t
 t = root (2 * ((97) / (9.8)))
 t = 4.449260429
 89% of your fall:
 0.89*97 = g * (t ^ 2/2)
 clearing t
 t = root (2 * ((0.89 * 97) / (9.8)))
 t = 4.197423894
 11% of your fall
 t = 4.449260429-4.197423894
 t = 0.252

 To know the speed when the last 11% of your fall begins, you must first know how long it took you to get there:
 86.33 = g * (t ^ 2/2)
 Determining t:
 t = root (2 * ((86.33) / (9.8))) =  4.19742389 s
 Then, your speed will be:
 vf = (9.8) * (4.19742389) = 41.135 m / s

 Speed just before reaching the ground:
 The time will be:
 t = 0.252 +  4.197423894 =  4.449423894 s
 The speed is
 vf = (9.8) * (4.449423894) = 43.603 m / s

 answer
 (a) t = 0.252 s
 (b) 41,135 m / s
 (c) 43.603 m / s

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !