A physics book slides off a horizontal table top with a speed of 1.60m/s . It strikes the floor after a time of 0.440s . Ignore air resistance. the following are given: v(0)=1.60 t= 0.440s ok this is the question Find the height of the table top above the floor. Take free fall acceleration to be g=9.80m/s^2 . i need to use this formula: x(t) = x(0) + v(0)t + 1/2at^2 <-- height formula the thing is why is v(0)t equal to zero? isnt the v(0) given in the problem? i know the correct answer and i know how the book done it, but they dont explain why v(0)t is equal to zero. x(0) = 0 + 1/2(-9.8m/s^2)(.440s)^2 <--- i dont get why v(0) is zero
ok for that question the answer is 0.949m, so you dont have to do the calculations. im working on this question: Find the vertical component of the book's velocity just before the book reaches the floor. express in m/s wouldnt it just be distance/time? 0.949m/0.440 = 2.15 which is incorrect
Your book should have explained that it was using the vertical component of the v(0) vector which happens to be zero (can you think of why this is so? Hint: the table is flat). The quantity you came up with was the average velocity of the book through it's transit from the table to the ground. There is an acceleration due to gravity, and as you know, accelerations are changes in velocity, so you will need to use the relation v = at; where a will of course be 9.8 m /s^2
the answer is -4.321, but why v=at? do i need the formula.... Vfy = V(0)y - gt v = -(-9.8)(0.44) = 4.321 which is wrong, why is that?
I apologize for my lack of explanation. I got v=at from the kinematical relation: [tex]v_f = v_i + a\Delta t[/tex] vi is simply zero since book slides off horizontally (with no initial vertical velocity.) [tex] \Delta t [/tex] is simply the time of flight. You got the wrong answer because you added in another minus sign that didn't need to be there. Hope I helped!
You know, that would actually make an interesting problem to figure out! (NERD ALERT!!) Correct me if i'm wrong but wouldn't it depend on the velocity of the horizontal launch? Launching the book at a critical speed would ensure it staying upright upon landing. Any other speed and it will either flip over or land on its edge.
Yes, it would depend on the velocity at launch as well as whether the book edge is parallel to the edge of the table! This was inspired by my observation that if I ever knock my toast off the counter it seems always to land jelly side down!