Let's break down and solve the problem step-by-step: --- ## **Problem Restatement** Coal is discharged from a dump truck with an initial velocity \[ \vec{v}_ = 5\, \text{ft/s} \angle 50^\circ \] It falls onto a conveyor belt \( B \). We need to **determine the required velocity \( v_B \) of the belt** if the relative velocity with which the coal hits the belt is to be: **(a) vertical** **(b) as small as possible** --- ## **Step 1: Break Down Initial Velocity** The initial velocity has a magnitude of \( 5\, \text{ft/s} \) at \( 50^\circ \) above the horizontal. - Horizontal component: \( v_{x} = 5 \cos 50^\circ \) - Vertical component: \( v_{y} = 5 \sin 50^\circ \) \[ v_{x} = 5 \times .6428 \approx 3.21\, \text{ft/s} \] \[ v_{y} = 5 \times .766 \approx 3.83\, \text{ft/s} \] --- ## **Step 2: Relative Velocity upon Hitting the Belt** Let’s denote: - \( v_B \) = velocity of the belt (to the right) - \( v_x \) = horizontal velocity of coal just before hitting the belt - \( v_y \) = vertical velocity of coal just before hitting the belt The **relative velocity** of the coal to the belt is: \[ \vec{v}_{rel} = (v_x - v_B)\ \hat{i} + v_y\ \hat{j} \] --- ## **Step 3: (a) Relative Velocity is Vertical** For the relative velocity to be vertical, its horizontal component must be **zero**: \[ v_x - v_B = \implies v_B = v_x \] The coal is under gravity, so as it falls, its horizontal velocity remains **constant** (no air resistance): \[ v_x = v_{x} = 3.21~\text{ft/s} \] **Final Answer (a):** \[ \boxed{v_B = 3.21~\text{ft/s}} \] --- ## **Step 4: (b) Relative Velocity is as Small as Possible** The relative velocity is minimized when the **magnitude** of the vector \((v_x - v_B)\ \hat{i} + v_y\ \hat{j}\) is minimized. The **minimum possible value** occurs when the relative velocity vector is **entirely vertical** (since the horizontal component can be canceled out by the belt's speed). Thus, **the answer is the same as part (a)**: \[ \boxed{v_B = 3.21~\text{ft/s}} \] **If you wanted the numerical value of the minimum relative velocity magnitude:** \[ v_{rel,\,min} = |v_y| \quad \text{(since horizontal part is zero)} \] But \(v_y\) will depend on the height from which the coal falls, which is not given, so this part cannot be further specified with the given information. --- ## **Summary Table** | Part | Required belt velocity \( v_B \) (ft/s) | |------|:--------------------------------------:| | (a) | 3.21 | | (b) | 3.21 | --- ### **Final Answers** **(a) \( v_B = 3.21~\text{ft/s} \) (to make the relative velocity vertical)** **(b) \( v_B = 3.21~\text{ft/s} \) (to make the relative velocity as small as possible)** --- Let me know if you need an explanation for the vector diagram or further details!