Fundamental of Physics 10th Edition Problem 52

[kmshowalter]

Homework Statement

A bolt is dropped from a bridge under construction, falling 83 m to the valley below the bridge. (a) How much time does it take to pass through the last 20 % of its fall? What is its speed (b) when it begins that last 20 % of its fall and (c) just before it reaches the ground?

Homework Equations

The Attempt at a Solution

For part (a) I got the correct answer of .436 seconds by doing 83/4.9=16.94, sqrt16.94=4.12
83 x .20 = 16.6
83-16.6=66.4
66.4/4.9=13.55
sqrt13.55=3.68
4.12-3.68=.436

That answer was correct but I need parts (b) and (c)

 

[SteamKing]

Knowing the time something is falling under gravity, how do you calculate the velocity? You've learned one equation of kinematics; there's a couple more equations you need to learn as well.

 

[kmshowalter]

I'm sorry but that really doesn't help my understanding of how to do it.

 

[SteamKing]

It means there are some other equations that you need to learn:

http://mathematicsi.com/deriving-suvat-equations-motion/

 

[HalfLostAndSearching]

as well.

For part (b), we can use the equation v = gt, where v is the final velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. We know that the final velocity when the bolt begins the last 20% of its fall will be the same as the initial velocity of the last 20% of its fall. So, using the calculated time from part (a) of 0.436 seconds, we can plug it into the equation to get: v = (9.8 m/s^2)(0.436 s) = 4.28 m/s. Therefore, the speed at the beginning of the last 20% of the fall is 4.28 m/s.

For part (c), we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. We know that the displacement for the last 20% of the fall is 16.6 m (as calculated in part (a)). We also know that the initial velocity is 4.28 m/s (as calculated in part (b)). And the acceleration is still 9.8 m/s^2. Plugging these values into the equation, we get: v^2 = (4.28 m/s)^2 + 2(9.8 m/s^2)(16.6 m) = 182.48 m^2/s^2. Taking the square root of both sides, we get: v = 13.5 m/s. Therefore, the speed just before the bolt reaches the ground is 13.5 m/s.