What is the horizontal distance between a tilted conveyer belt and a pipe?

[burton95]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

3 + 6.0sin15t -4.9t²

t= .956797

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

not it

 

[TSny]

3 + 6.0sin15t -4.9t²

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

 

[burton95]

Hi.

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

0 = 3 + 6.0sin15t -4.9t2

the sign on initial velocity is +

 

[TSny]

the sign on initial velocity is +

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

 

[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

 

[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.

 

[burton95]

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

Because using -g turns my squared term positive

 

[burton95]

Nevermind. Thx u were a great help