A bolt 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