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
Phyiscs - bobpursley, Saturday, July 11, 2009 at 8:00pm
Vo is zero, starting at the top.
T= sqrt(2x/g) where g is the component of gravity down the plane 9.8sin22
Phyiscs - drwls, Saturday, July 11, 2009 at 8:12pm
If the plane is frictionless, the block will rise a vertical height h so thet mgh equals the initial kinetic energy (1/2) mV^2. Thus h = V^2/(2g) = 0.81 m. The distance x that the block mov es up the track is h/sin22 = 2.2 m
So your answer is correct.
The block decelerates at a rate
a = g sin 22 = 3.67 m/s^2.
The block reaches maximum height when
t = V/a = 4/3.67 = 1.1 s
It takes the same length of time to come back down. The total e;apsed ti9me to return is thus 2.2 s
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