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pstfleur
Sep1-08, 11:54 PM
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+Vot-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 Im even headed in the right direction.

lax1113
Sep2-08, 12:24 AM
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=VoT+.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!

pstfleur
Sep2-08, 05:36 PM
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=VoT+.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

grosenblatt
Sep2-08, 05:59 PM
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 in to 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