# Free fall acceleration

1. Sep 18, 2007

### suxatphysix

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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.

$$\Delta$$t=-vi +/- $$\sqrt{vi^{2}+2ay}$$ / a
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 18, 2007

### learningphysics

That's a good approach. What answers did you get?

3. Sep 18, 2007

### suxatphysix

got 8.20 for part a seconds but it was wrong. so i stopped at part a

4. Sep 19, 2007

### learningphysics

Can you show what you did to get that?

5. Sep 19, 2007

### suxatphysix

vf$$^{2}$$=0 + 2(-9.8m/s^2)(-68.6m)
vf= 36.69m/s

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

=8.20s

6. Sep 19, 2007

### learningphysics

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: Sep 19, 2007
7. Sep 19, 2007

### suxatphysix

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: Sep 19, 2007
8. Sep 19, 2007

### learningphysics

9. Sep 19, 2007

### suxatphysix

$$\Delta$$t=-vi +/- $$\sqrt{vi^{2}+2ay}$$ / a

10. Sep 19, 2007

### learningphysics

Ah... I think I see now.

Here you should have taken vf as -36.69m/s, not +36.69m/s. ie the negative square root is what you want.

11. Sep 19, 2007

### learningphysics

Yes, that equation is correct.

12. Sep 19, 2007

### suxatphysix

omg it was right! thanks!

now to try and answer part b and c.

13. Sep 19, 2007

### learningphysics

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. Sep 19, 2007

### suxatphysix

i got both b and c.

b - 36.69
c - 43.85

thanks learningphysics for getting me started on that problem.
Much appreciation.

15. Sep 19, 2007

### learningphysics

cool! no prob. good job!