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
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???
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
thanks for helping me solve this problem
thanks
so the equation is
x = 2^-1 a t^2
were I know a is about 3.671 s^-1 m
thanks guys!
To find the time it takes for the block to return to its starting point, we can use the kinematic equation:
x = 0 = x₀ + v₀t + (1/2)at²
In this equation, x is the displacement (in this case, the block's vertical position), x₀ is the initial displacement (starting point), v₀ is the initial velocity, t is the time, and a is the acceleration.
Since the block goes up the plane and then returns to its starting point, its final displacement is 0. We also know the initial velocity v₀ and the acceleration a due to gravity (which is -9.8 m/s²).
Rearranging the equation to solve for time (t), we have:
0 = x₀ + v₀t + (1/2)at²
This equation is quadratic in terms of t, so we can solve it using the quadratic formula:
t = (-v₀ ± √(v₀² - 2ax₀)) / a
In this case, the positive root will give us the time it takes for the block to return to its starting point, considering the initial velocity is upward.
Regarding your calculation for the distance the block travels up the plane, you correctly used the equation:
v² = v₀² + 2a(x - x₀)
To rearrange for the displacement (x - x₀), you have:
x - x₀ = (v² - v₀²) / (2a)
By substituting the given values into the equation, you got the correct answer of 2.2 m.
It's important to resolve the initial velocity (v₀) into its components if the given value is not already in the desired direction. In your case, since the block is moving up the 22-degree plane, you need to resolve the initial velocity into its horizontal and vertical components. The vertical component can be determined by multiplying the initial velocity by the sine of the angle (sin(22°)), and that would indeed account for the effect of gravity on acceleration.