Physics Help Forum Falling object

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 10th 2009, 04:50 AM #21 Banned   Join Date: Aug 2009 Location: UK Posts: 240 OOps !!! I made a mistake with the decimal fraction, it should be 0.404694. So 1 sec / ( 1-(sq-rt ( 1-0.404694)) = 4.3775 secs total time = (1/2)at^2 = 0.5 x 9.8 x 4.3775^2 = 93.89 m my mistake, your answer is correct. Whenever this sort of question shows it's face again i'll memorize it using the following: 38m = 38 m/s + (1/2)g = v (final velocity) = 38 + 4.9 v = gt t= v/g = 42.9 m/s / 9.8 = 4.3775 secs (1/2)gt^2 = 93.89 m Last edited by Paul46; Oct 10th 2009 at 06:47 AM.
 Oct 10th 2009, 10:49 PM #22 Physics Team   Join Date: Feb 2009 Posts: 1,425 I am kind of relieved the answer is correct, but i dont understand this bit So 1 sec / (1-(sq-rt (1-0.35681)) = 5.05 secs total time falling I know the 0.35 bit is wrong from your later post, but how do you come to this ratio at all ? 1 sec / (1-(sq-rt (1-0.35681)) Last edited by physicsquest; Oct 10th 2009 at 11:24 PM.
 Oct 11th 2009, 02:44 AM #23 Banned   Join Date: Aug 2009 Location: UK Posts: 240 physicsquest, It's just a ratio to confirm the total time. A stone is dropped from a cliff in the final second the stone travels 67m, find the the total displacement. 67m = v= 67 m/s + (1/2)g = 71.9 m/s v=gt t=v/g = 7.3367 secs (1/2)gt^2 = 263.755m total displacement To confirm is the answer was correct i would use: 67m = 0.254023 decimal fraction of total displacement. final 1 sec/(1-sq-rt(1-0.254023)) = 7.3367 secs But since now i know that the final velocity (v) will be displacement (67m) + (1/2g) i no longer need to confirm the answer using that formula. It may seem abit odd, but it's the way i like to drum it into myself. Once i found your answer was correct i started breaking the calculations down to find a simple calculation of the answer, because i couldn't work the answer out, when i broke the calculations down i found the easy way to find (v) was displacement + (1/2)g then do the above calculations. I hope this answers your question.
 Oct 11th 2009, 09:06 AM #24 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,576 Do I understand both of you right now, i.e. all three of us agree on the value of H? If so then I'm quite happy!
 Oct 11th 2009, 12:25 PM #25 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 132 Falling Object Using the last second S = 38 U is unknown, A is Gravity and T is one. S = ut + 0.5At^2 U = 33.1 M / S at the beginning of the final second V = U + AT U = 0, A is Gravity T is unknown to find the time up to point U = 33.1 T = 3.37 Seconds T for the total journey is 4.37 seconds. Substituting back into S = Ut + 0.5AT^2 with U = 0 A Gravity T = 4.37 gives H = 93.89. Give or take a few millimeters depending on the flavour of G (9.8 used here) G of 9.81 gives H as 93.84 This seems to confirm the number already calculated before.
 Oct 12th 2009, 12:09 PM #26 Banned   Join Date: Aug 2009 Location: UK Posts: 240 Just out of interest guys, I've been thinking is there a way to solve the above sort of question using longer time durations. Tapping away on my calculator for a few hours i found how to do it. A stone is dropped from a cliff, in the last 3 secs the stone travels 99m, find the total displacement. u = initial velocity v=final velocity a=acceleration (9.8 m/s^2) t=time t1=given time (3) s=displacement s1= given displacement (99) This is what i did to find it: v = s1/t1 + 3a/2 = 33 + 14.7 = 47.7 m/s t= v/a = 4.867 s (3 d.p) s= ut + (1/2)at^2 = 116.07 (2 d.p) Just to confirm: 4.867 s - 3 s = 1.867 s (1/2)at^2 = 17.07 17.07 + 99 = 116.07 Last edited by Paul46; Oct 12th 2009 at 12:29 PM.

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