A box moves down a conveyor belt the is angled at 15 degrees, at a rate of 6m/s. The box then enters a dump off that is 3m below the end of the belt, what is the horizontal distance between the dump off and the belt?

I have no idea where to start with this...I know you have x and y components of something but I don't know what to do...any help in the right direction would be appreciated.

To solve this problem, we can break down the motion of the box into its horizontal (x) and vertical (y) components.

First, let's find the time it takes for the box to reach the dump off. We know the vertical displacement (y) is -3m (negative because the box is moving downward) and the initial vertical velocity (Viy) is 6 sin(15°). We can use the equation of motion for vertical motion:

y = Vit + (1/2)gt^2

Substituting the known values:

-3 = (6 sin(15°))t + (1/2)(9.8)t^2

Simplifying this equation, we get a quadratic equation:

0.5(9.8)t^2 + 6 sin(15°)t - 3 = 0

Now, we can solve this equation to find the time (t) it takes for the box to reach the dump off. You can solve this equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

where a = 0.5(9.8), b = 6 sin(15°), and c = -3.

Once you find the value(s) of t, you can substitute it back into the equation for horizontal motion:

x = Vix * t

where Vix is the initial horizontal velocity, which is 6 cos(15°).

By substituting the calculated values, you can find the horizontal distance (x) between the dump off and the belt.