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

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SUMMARY

The discussion focuses on calculating the total time it takes for a wrecking ball to fall to the ground after its cable breaks. The time to fall halfway is established as 1.2 seconds. Using the equations of motion, specifically ΔY = V0t + 0.5gt² and v² = V0² + 2gD, participants derive that the total time to fall is approximately 2.88 seconds. The acceleration due to gravity (g) is consistently referenced as 9.81 m/s² throughout the calculations.

PREREQUISITES
  • Understanding of basic physics concepts, particularly kinematics
  • Familiarity with equations of motion, including ΔY = V0t + 0.5gt²
  • Knowledge of gravitational acceleration (g) as 9.81 m/s²
  • Ability to manipulate algebraic equations for solving time and distance
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to calculate free fall time using different initial conditions
  • Explore the effects of air resistance on falling objects
  • Investigate real-world applications of projectile motion in engineering
USEFUL FOR

Students studying physics, educators teaching kinematics, and engineers involved in structural demolition or related fields will benefit from this discussion.

pstfleur
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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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