A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of \(14.0^{\circ}\) from the horizontal and the sand is moved without slipping at the rate of \(7.00 \mathrm{~m} / \mathrm{s}\). The sand is collected in a big drum \(3.00 \mathrm{~m}\) below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.

We can use the following kinematic equation to find the time it takes for the sand to fall from the end of the conveyor belt to the middle of the drum: $$y = v_{0y} t + \frac{1}{2} a t^2$$ where \(y\) is the vertical displacement, \(v_{0y}\) is the initial vertical velocity, \(t\) is the time, and \(a\) is the vertical acceleration due to gravity. Since the sand falls freely from the end of the conveyor belt, we can assume that the initial vertical velocity \(v_{0y} = 0\), and the acceleration due to gravity \(a = -9.81 \mathrm{~m/s^2}\). We can rearrange the equation to solve for \(t\): $$t = \sqrt{\frac{2y}{a}}$$ Plug in the given value for \(y = 3.00 \mathrm{~m}\) and \(a = -9.81 \mathrm{~m/s^2}\): $$t = \sqrt{\frac{2 \times 3.00}{-9.81}} \approx 0.779 \mathrm{~s}$$

We can calculate the horizontal component of the sand's velocity by using the angle of the conveyor belt and the given speed: $$v_x = v \cos{\theta}$$ where \(v\) is the speed of the sand on the conveyor belt and \(\theta\) is the angle between the conveyor belt and the horizontal. Plug in the given values for \(v = 7.00 \mathrm{~m/s}\) and \(\theta = 14.0^{\circ}\) (convert to radians): $$v_x = 7.00 \cos{14.0^\circ} = 7.00 \cos{0.244} \approx 6.79 \mathrm{~m/s}$$

We now have the time it takes for the sand to fall and the horizontal component of its velocity. To find the horizontal distance between the end of the conveyor belt and the middle of the collecting drum, we can use the following equation: $$x = v_x t$$ Plug in the values we calculated for \(v_x = 6.79 \mathrm{~m/s}\) and \(t = 0.779 \mathrm{~s}\): $$x = 6.79 \times 0.779 \approx 5.29 \mathrm{~m}$$ Therefore, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is approximately 5.29 meters.