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Ok I asked this question before

A lbock is given an intial speed of 4.0 s^-1 m up the 22 degree plane

How far up the plane will it go
ok I got this to be something like 2.2 about i understand this

i need help on this
how much time elapses before it returns to tis starting point? Ignore friction.
ok I need to know what formula to use to rearange for time

i thought it was this

x = 2^-1 a t^2 + Vo t + Xo

if i'm right then i guess I do not know how to rearange this for time because ther's a t^2 and t so could show me how to rearange this for time

ok and to find the distance i did this formula

V^2 = Vo^2 + 2a(X-Xo)
rearanged for X
x = (2a)^-1 (-Vo)^2
and if i did that correctly
I got 2.2 m so is this a valid way to doing this also

by the way I was told from

drwls this

You could also solve
y = V t - (a/2) t^2 = 0
to get the answer. Ignore the t=0 solution.
t = 8/a = 2.2

to solve for time

I just want to confirm that you had to resolve the Vo into it's componetns right to get Vo in the y direction correct and used gravity for accelration???

  • Physics -

    hmmm aparently that's not how you do it
    hmmm could you please show me how to get the time step by step

    don't you have to use this equation
    X = 2^-1 a t^2 + Vo t + Xo
    write it in the second dimension
    Y = 2^-1 g t^2 + Vo t + Yo

    and sense there is no Yo
    I get this

    Y = 2^-1 g t^2 + Vo t
    still not enoguh to solve for time
    so I don't get it

    for the y component of the inital velocity I got about 1.498 s^-1 m
    I do belive there is one...

  • Physics -

    If you start at the top, there is no initial velocity. Find the time downward, then double if for the entire trip.

  • Physics -

    oh... lol thanka


    lol ok um so then I have this equation

    y = 2^-1 g t^2
    ok I get it
    however I do not know y
    how do I get y

    thanks for helping me solve this problem

  • i get it now -

    so the equation is
    x = 2^-1 a t^2
    were I know a is about 3.671 s^-1 m
    thanks guys!

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