Free fall acceleration of a bolt

  • #1

Homework Statement



A bolt is dropped from a bridge under construction, falling 98 m to the valley below the bridge.

(a) In how much time does it pass through the last 30% of its fall?


(b) What is its speed when it begins that last 30% of its fall?


(c) What is its speed when it reaches the valley beneath the bridge?


Homework Equations





The Attempt at a Solution


Tried splitting the problem into 2 parts by multiplying 98 x .3 to solve for the final velocity for the first 70% of the fall, then using that velocity as the initial velocity for the 30% of the fall and the other 30%(29.4m) of the trip.

[tex]\Delta[/tex]t=-vi +/- [tex]\sqrt{vi^{2}+2ay}[/tex] / a

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
6

Homework Statement



A bolt is dropped from a bridge under construction, falling 98 m to the valley below the bridge.

(a) In how much time does it pass through the last 30% of its fall?


(b) What is its speed when it begins that last 30% of its fall?


(c) What is its speed when it reaches the valley beneath the bridge?


Homework Equations





The Attempt at a Solution


Tried splitting the problem into 2 parts by multiplying 98 x .3 to solve for the final velocity for the first 70% of the fall, then using that velocity as the initial velocity for the 30% of the fall and the other 30%(29.4m) of the trip.
That's a good approach. What answers did you get?
 
  • #3
got 8.20 for part a seconds but it was wrong. so i stopped at part a
 
  • #4
learningphysics
Homework Helper
4,099
6
got 8.20 for part a seconds but it was wrong. so i stopped at part a
Can you show what you did to get that?
 
  • #5
vf[tex]^{2}[/tex]=0 + 2(-9.8m/s^2)(-68.6m)
vf= 36.69m/s

[tex]\Delta[/tex]t = -36.69m/s +/-[tex]\sqrt{36.69m/s^2 + 2(-9.8m/s^2)(-29.4) }[/tex] /-9.81

=8.20s
 
  • #6
learningphysics
Homework Helper
4,099
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vf[tex]^{2}[/tex]=0 + 2(-9.8m/s^2)(-68.6m)
vf= 36.69m/s

[tex]\Delta[/tex]t = -36.69m/s +/-[tex]\sqrt{36.69m/s^2 + 2(-9.8m/s^2)(-29.4) }[/tex] /-9.81

=8.20s
Everything looks right except, your denominator in your quadratic solution... I think it should be 9.81, not -9.81.... so you'll chose the positive numerator instead of the negative one...
 
Last edited:
  • #7
why 9.81 when gravity is always negative?


that means the time would actually be 0.723? i dont know if that sounds logical
 
Last edited:
  • #8
learningphysics
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why 9.81 when gravity is always negative?
Can you write out the quadratic equation you had?
 
  • #9
[tex]\Delta[/tex]t=-vi +/- [tex]\sqrt{vi^{2}+2ay}[/tex] / a
 
  • #10
learningphysics
Homework Helper
4,099
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Ah... I think I see now.

vf=0 + 2(-9.8m/s^2)(-68.6m)
vf= 36.69m/s
Here you should have taken vf as -36.69m/s, not +36.69m/s. ie the negative square root is what you want.
 
  • #11
learningphysics
Homework Helper
4,099
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[tex]\Delta[/tex]t=-vi +/- [tex]\sqrt{vi^{2}+2ay}[/tex] / a
Yes, that equation is correct.
 
  • #12
omg it was right! thanks!

now to try and answer part b and c.
 
  • #13
learningphysics
Homework Helper
4,099
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omg it was right! thanks!

now to try and answer part b and c.
cool! another way to solve that problem that I just saw... you can get the total time of the drop. and subtract the time of the first 70%

So solving (1/2)gt^2 = 98, gives t = 4.4699s. And the solve the time for the first 70%. (1/2)gt^2 = 0.7*98, gives t = 3.7397s, so the difference is the time you need 0.730s.
 
  • #14
i got both b and c.

b - 36.69
c - 43.85


thanks learningphysics for getting me started on that problem.
Much appreciation.
 
  • #15
learningphysics
Homework Helper
4,099
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i got both b and c.

b - 36.69
c - 43.85


thanks learningphysics for getting me started on that problem.
Much appreciation.
cool! no prob. good job!
 

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