A bolt is dropped from a bridge under construction, falling 94 m to the valley below the bridge. (a) how much time does it take to pass through the last 26 % of its fall? what is its speed (b) when it begins that last 26 % of its fall and (c) just before it reaches the ground?

Refer to the diagram shown below.

When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.

The last 26% of its fall is at a height of 0.26*94 = 24.4 m.

At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.

The time, t, for the bolt to fall a known distance obeys the equation

s = Vt + (1/2) gt²,

where

s = 69.56 m, vertical distance traveled, and

g = acceleration due to gravity.

Therefore

69.56 = 0 + (1/2) * 9.8*t²

t² = (69.56*2) / 9.8

t = 3.7677 s

The total time, T, to fall 94 m is given by

94 = (1/2) * 9.8*T^2

T² = 19.1837

T = 4.38 s

The time taken to pass through the last 26% of its fall is

T - t = 4.38 - 3.7677 = 0.6122 s

The speed after falling 69.56 m is given by

V₁ = 0 + g*t = 36.9235 m/s

The speed with which the bolt strikes the ground is given by

V₂ = 0 + g*T = 9.8*4.38 = 42.924 m/s

Answer:

(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.

(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).

(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).