Let the cat take t sec to fall 1.6 m By s = ut + 1/2gt^2 1.6 = 0 + 1/2 x 9.81 x t^2 t =?0.33 = 0.57 sec By R = Ux x t 2.1 = you x 0.57 you = 3.68 m/s.
YOU aren't missing anything. I have written online answer systems for people and they are both difficult and stupid. I.
E How many significant figures do they expect? What if a student answers in a different way from that expected? And this doesn't even start to address the fact that the machines are simply WRONG sometimes.
Both of the answers confirm that your answer to the question given is correct. As both the height and the distance are to two significant figures then there should only be two in the answer. I.
E 2.2 m/s is correct for the question you have given us. So either the question is wrong or THEIR answer is wrong. When you do things right you MUST have confidence in your result.
Eventually someone will need you to give answers and when there is no one else to check your work you must be confident enough to say "THIS is done correctly. It will work".
Use the equation t =?2y/g the entire expression is square rooted and y represents the height of the table?(2 x 1.4m)/9.81 m/s² = 0.53 seconds 1.2m/0.53 seconds = 2.3 m/s.
I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.