Ok I asked this question before

Question

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
thanks

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???

Answer 1

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...

Answer 2

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

Answer 3

oh... lol thanka

Answer 4

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

thanks for helping me solve this problem

Answer 5

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