# Incline solving for xf

1. Feb 2, 2013

### burton95

1. The problem statement, all variables and given/known data

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15°. The sand enters a pipe 3.0 m below the end of the conveyer belt, as shown in the figure below. What is the horizontal distance d between the conveyer belt and the pipe?

Xi = 0 m
Ti = 0 s
Viy = 6.0 sin θ
Yi = 3 m
Vix = 6.0 cos theta
Xf = ???
Tf = t
Vf =
Yf = 0

2. Relevant equations

Yf = Yi + Viy(t) - 1/2g(t)2
Xf = Xi + Vix(t)

3. The attempt at a solution

3 + 6.0sin15t -4.9t2

t= .956797

Xf = 0 + 6*cos15*.956797 = 5.54 meters

not it

2. Feb 2, 2013

### TSny

Did you mean to write an equation here? What is the sign of the initial y-component of velocity?

3. Feb 2, 2013

### burton95

Hi.

Yf = Yi + Viy(t) - 1/2g(t)2

0 = 3 + 6.0sin15t -4.9t2

the sign on initial velocity is +

4. Feb 2, 2013

### TSny

The problem says that the sand moves down the conveyor. So, the y-component of velocity should be down.

5. Feb 2, 2013

### burton95

I thought that the direction was accounted for in setting Yi = 3 and Yf = 0.

so my proper equation 0 = 3 - 6.0sin15t -4.9t2

6. Feb 2, 2013

### TSny

That looks good. Note that you took yf =0 and yi = 3 which is correct for taking upward as positive. You also used -9.8 m/s^2 for the acceleration which agrees with taking upward as positive since the acceleration is downward. Likewise, the y component of the initial velocity is downward, so it should be negative when taking upward as positive.

7. Feb 3, 2013

### burton95

Is this correct
Yi = Yo + Voy (Tf - Ti) - (1/2)(g)(Tf-Ti)^2

Because using -g turns my squared term positive

8. Feb 3, 2013

### burton95

Nevermind. Thx u were a great help