Question

A bolt is dropped from a bridge under construction, falling 95 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?

Answer 1

Answer on Question #51370, Physics, Mechanics | Kinematics | Dynamics

Task: A bolt is dropped from a bridge under construction, falling 95 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?

Solution:

So 26% of 95 is 24.7m. But we need to know total time for the free-fall first.

So using our trusty old acceleration and distance equation where initial distance and initial velocity are both 0.

$$h = V_0 \cdot V_0 t + g t^2 / 2$$ $$95\,\mathrm{m} = 0 + 0 + 9.8 \cdot t^2 / 2$$ $$95\,\mathrm{m} = 4.9 t^2$$ $$19.39 = t^2$$

t=4.403 seconds for it to fall the whole 95 m

Now the last 26% of its fall means it has already fallen 95 - (0.26*95) = 70.3m

So we want to know the time it takes for the bolt to fall 70.3m now.

With initial velocity and distance still equal to zero

$$70.3\,\mathrm{m} = 4.9 t^2$$

t for 72m is 3.787s

So 4.403 - 3.787 = 0.616 s to fall the rest of the last 26% of its fall

b) Well since it has traveled 70.3 m in 3.787 seconds, then

$$v = 70.3 / 3.787$$ $$v = 18.56 \,\mathrm{m/s}$$

c) 95 m in 4.403s means

$$v = 95 / 4.403$$ $$v = 21.58 \,\mathrm{m/s}$$

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