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.
:You have to figure the time it took to fall 3m. It had an initial vertical velocity downward of -6sin15.
h=vi*t -4.9t^2
solve for t.
Then, you know the intial horizontal velocity is 6cos15, and the time, so what distance did it travel horizontally?
You can't solve for t from that though can you? If you use quadratic equation its a negative sign under the root and you can't take the root of that. I'm confused:S
To solve this problem, we can break down the motion of the box into its horizontal and vertical components.
First, let's consider the horizontal component. The box is moving at a constant rate of 6 m/s along the conveyor belt. The time it takes for the box to reach the dump off will be the same as the time it takes for the box to fall 3 m vertically.
To find the time it takes for the box to fall, we can use the formula for the time it takes for an object to fall freely due to gravity:
t = sqrt(2h/g)
where
t = time (in seconds)
h = vertical distance (in meters)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values, we have:
t = sqrt(2 * 3 m / 9.8 m/s^2)
= sqrt(6 / 9.8)
≈ 0.78 s (rounded to two decimal places)
Now that we know the time it takes for the box to fall, we can calculate the horizontal distance it moves in that time. We can use the formula:
distance = speed * time
where
distance = horizontal distance (in meters)
speed = horizontal speed (in meters per second)
time = time (in seconds)
Substituting the values, we have:
distance = 6 m/s * 0.78 s
≈ 4.68 m (rounded to two decimal places)
Therefore, the horizontal distance between the dump off and the end of the belt is approximately 4.68 meters.