- #1
rexexdesign
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Homework Statement
A cat drops from a shelf 3.7ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12cm.
Part A: Calculate his speed when he first touches the floor (ignore air resistance).
Part B: Calculate how long it takes him to stop.
Part C: Calculate his acceleration (assumed constant) while he is stopping, in m/s2.
Part D: Calculate his acceleration (assumed constant) while he is stopping, in g’s.
Homework Equations
For Part A: V2y = V2y0+2ay(y-y0)
I am stuck at part B, I know I need to find the time that it takes to slow down, right? This is my first week of Physics and I am not fluent in using the formulas or knowing what Vyo always is and what to use for acceleration. I tried to use this formula: y = yo + vo t + ½ a t2
The Attempt at a Solution
Part A (correct) 0 + (2*32.2ft/s2) * (3.7ft-0) = 4.7m/s (which I entered in my HW and it's correct)
Part B When I use Vo = 4.7m/s and a=9.8m/s2 and V=0 (because the final velocity would be 0 when the cat stopped completely, right?) I get 0.16s and the program said it was wrong. I also got another result by changing the equation of 0.11s which was wrong as well. Finally I looked in my physics book and found the SAME problem, but the cat jumped from 4.0ft instead of 3.7ft. They got 4.9m/s velocity as the first answer, and 0.049s as the time to stop... so I though well it must be 0.047s for me then... wrong. Also didn't work. I only have 3 attempts left on this question and I am stuck. The book got an acceleration of 100m/s2 for part C and 10g for part D, which even confused me more. How can it have such a high acceleration?
Any help would be much appreciated, the HW is due tomorrow and I was up till 1 AM last night trying to figure this out, no problem in the book deals with a deceleration within a certain distance like this.
I am sure this is easy to most of you, but I am just getting my feet wet with Physics!
Thanks!