Loose bolt falling from elevator

[jrk613]

Homework Statement

A bolt comes loose from the bottom of an elevator that is moving upward at a speed of 6.88 m/s. The bolt reaches the bottom of the elevator shaft in 3.90 s. How high up was the elevator when the bolt came loose?

Homework Equations

d= v(initial)t + .5at^2

The Attempt at a Solution

d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong? Thanks

 

[rl.bhat]

1.


d= v(initial)t + .5at^2

The Attempt at a Solution

d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong? Thanks

In the problem the displacement ang acceleration due to gravity are in the same direction, but the velocity is in the opposite direction. Accordingly use the proper signs of these quantities.

 

[jrk613]

Changing the signs for the first half of the equation i get either + or - 26.832... for the second half i get + or - 75.21. I've tried combining all 4 combinations and all 4 are said to be wrong... any more ideas? thanks again.

 

[rl.bhat]

Try this
d = - V(initial)*t + 0.5*g*t^2

 

[Jobrag]

Remember that before the bolt starts to fall it will carry on up till gravity brings it to a halt.

 

[jrk613]

eh, no luck, thanks anyways folks... i'll find out the answer today

 

[Doc Al]

d= 6.88(3.9) + .5(-9.89)(3.9)^2

I got the answer -48.38 but the distance is wrong... anyone see where I went wrong?

Are you measuring time from when the elevator started rising or from the moment that the bolt began to fall?

If the latter, then what you have is fine except for a sign error. But perhaps the problem is the more interesting one that measures time from the moment that the elevator begins moving upward. In which case you must treat it in two parts: (1) the elevator rises to height h, and (2) the bolt falls from height h to the bottom of the shaft.