Posted by **Physics** on Saturday, July 11, 2009 at 7:39pm.

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

- 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

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