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

Answer 1

Vo is zero, starting at the top.

T= sqrt(2x/g) where g is the component of gravity down the plane 9.8sin22

Answer 2

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