How Long Does It Take for a Wrecking Ball to Fall to the Ground?

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Homework Help Overview

The discussion revolves around the physics of a wrecking ball falling from rest after its cable breaks. Participants are exploring the time it takes for the ball to fall to the ground, given that it takes 1.2 seconds to fall halfway.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations of motion, including those involving initial velocity, acceleration due to gravity, and time. Some express uncertainty about how to relate their calculations to the overall problem.

Discussion Status

Several participants have offered different equations and approaches to tackle the problem. There is a mix of interpretations regarding the variables used, and some participants are questioning the assumptions made in their calculations. Guidance has been provided on how to approach the problem, but no consensus has been reached on a definitive solution.

Contextual Notes

Some participants mention differences in variable notation from their previous learning experiences, which may affect their understanding of the equations being discussed. There is also a focus on calculating the distance fallen and the implications of using different equations of motion.

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1. A wrecking ball is hanging at rest from a crane when suddenly the cable breaks. The time it takes for the ball to fall halfway to the ground is 1.2 seconds. Find the time it takes for the ball to fall from rest all the way to ground
2. Vf=Vo-gt, Vf^2=Vo^2-2g(delta y), Yf=Yo+volt-1/2gt^2
3. I first found the velocity of the ball traveling to the halfway mark.
Vf=0m/s-9.8(1.2)=-11.76.. I'm not quite sure how to relate that to the actual question. I wonder if I am even headed in the right direction.
 
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Hello
Please excuse me because in my physics class we learned with a different set of variables, which seems stupid because on PF i have yet to see anyone use the ones we have haha. Anyway, the equation needed to solve this would be.

∆Y=volt+.5At2 (i believe A is called g in your equations)

In words this reads... change in y is equal to the initial velocity multipled by time, plus half the acceleration multiplied by the time squared.

So, what you should look to find out, is the change in y of the half of it, which took 1.2 seconds. after you get that, I am sure that you will be well on your way. To check if you do the problem right, the answer i got is sqare root of 2.88 seconds2

Good luck!
 
lax1113 said:
Hello
Please excuse me because in my physics class we learned with a different set of variables, which seems stupid because on PF i have yet to see anyone use the ones we have haha. Anyway, the equation needed to solve this would be.

∆Y=volt+.5At2 (i believe A is called g in your equations)

In words this reads... change in y is equal to the initial velocity multipled by time, plus half the acceleration multiplied by the time squared.

So, what you should look to find out, is the change in y of the half of it, which took 1.2 seconds. after you get that, I am sure that you will be well on your way. To check if you do the problem right, the answer i got is sqare root of 2.88 seconds2

Good luck!

Thanks, That helped extremely in putting me in the right direction
 
well, downward acceleration on Earth (or g) is always a constant 9.81 m/s2 so ll you have to do is plug into the equation
so just plug that into the equation D = D0 + V0t +(1/2)at2 to find the half of the distance, then double to find the total distance.
then we change the equation v2 = v02 + 2aD so that it can solve for v, v0 is 0 and 1 = g so all we do is remove that, switch a to g, and take the squareroute of both sides snd we get. V = sqrt(2gd), plug in and solve. the use that value as v in the equation v = at (a = g again) and solve for total time
 
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