posted by Physics on .
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
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???
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...
If you start at the top, there is no initial velocity. Find the time downward, then double if for the entire trip.
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
so the equation is
x = 2^-1 a t^2
were I know a is about 3.671 s^-1 m