# Free fall acceleration of a bolt

## 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?

## 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

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
learningphysics
Homework Helper

## 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?

## 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?

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

learningphysics
Homework Helper
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?

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

learningphysics
Homework Helper
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
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:
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:
learningphysics
Homework Helper
why 9.81 when gravity is always negative?
Can you write out the quadratic equation you had?

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

learningphysics
Homework Helper
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.

learningphysics
Homework Helper
$$\Delta$$t=-vi +/- $$\sqrt{vi^{2}+2ay}$$ / a
Yes, that equation is correct.

omg it was right! thanks!

now to try and answer part b and c.

learningphysics
Homework Helper
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.

i got both b and c.

b - 36.69
c - 43.85

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

learningphysics
Homework Helper
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!