Let's address **part 1** and **part 2** step by step: --- 1. **Write the parametric equations of motion for the projectile** ### **Step 1: Analyze Initial Velocities** - Platform moves at \( u = 8\, \text{m/s} \) (horizontal, to the right). - Projectile fired at \( v_ = 30\, \text{m/s} \) at \( \theta = 40^\circ \) (relative to platform). **Relative to ground:** - Horizontal velocity: \( v_{x} = u + v_ \cos\theta \) - Vertical velocity: \( v_{y} = v_ \sin\theta \) **Explanation:** The horizontal velocity seen by an observer on the ground is the sum of the platform's velocity and the projectile's component along the horizontal (from the platform). --- ### **Step 2: Write Parametric Equations** **Horizontal:** \[ x(t) = (u + v_\cos\theta) t \] **Vertical:** \[ y(t) = v_\sin\theta \cdot t - \frac{1}{2}gt^2 \] **Explanation:** The position in each direction is given by the initial velocity in that direction times time, and for vertical motion, minus the effect of gravity. --- ## Final Answer for 1: \[ \boxed{ \begin{align*} x(t) &= (u + v_\cos\theta)\, t \\ y(t) &= v_\sin\theta\, t - \frac{1}{2}gt^2 \end{align*} } \] --- --- ## 2. **Find the horizontal distance traveled when the projectile returns to ground level** ### **Step 1: Find Time of Flight** The projectile returns to ground when \( y(t) = \): \[ = v_\sin\theta \cdot t - \frac{1}{2}gt^2 \] \[ t \left( v_\sin\theta - \frac{1}{2}gt \right) = \] Ignoring \( t = \): \[ v_\sin\theta = \frac{1}{2}gt \implies t = \frac{2v_\sin\theta}{g} \] **Explanation:** This gives the total time the projectile spends in the air, considering only vertical motion. --- ### **Step 2: Calculate Horizontal Range** Plug \( t_{\text{flight}} \) into \( x(t) \): \[ R = x(t_{\text{flight}}) = (u + v_\cos\theta)\cdot \frac{2v_\sin\theta}{g} \] Given values: - \( u = 8\, \text{m/s} \) - \( v_ = 30\, \text{m/s} \) - \( \theta = 40^\circ \) - \( g = 9.8\, \text{m/s}^2 \) Calculate: - \( \sin 40^\circ \approx .6428 \) - \( \cos 40^\circ \approx .766 \) \[ R = [8 + 30 \cdot .766] \cdot \frac{2 \cdot 30 \cdot .6428}{9.8} \] \[ = [8 + 22.98] \cdot \frac{38.568}{9.8} \] \[ = 30.98 \cdot 3.936 \] \[ \approx 121.9\, \text{m} \] **Explanation:** This is the distance the projectile travels horizontally before hitting the ground, as seen from the ground frame. --- ## Final Answer for 2: \[ \boxed{R \approx 121.9\, \text{m}} \]

Let's continue with **part 3** and **part 4** step by step. --- ## 3. **Describe how the acceleration changes and explain why the existing mechanics apply** ### **Step 1: Analyze Acceleration** The projectile experiences two types of acceleration: - **Horizontal Acceleration**: \( a_x = 0 \) (constant horizontal velocity) - **Vertical Acceleration**: \( a_y = -g \) (due to gravity) **Explanation:** The horizontal motion is uniform (no horizontal acceleration) since the platform moves at a constant velocity. In contrast, the vertical motion is uniformly accelerated due to the constant pull of gravity acting downwards. --- ### **Step 2: Relate to Existing Mechanics** In projectile motion: - Horizontal motion is independent of vertical motion. - Vertical motion is influenced by gravity and results in a parabolic trajectory. **Explanation:** The principles of kinematics apply here because the horizontal and vertical motions can be analyzed separately, allowing us to use basic equations of motion for each direction. --- ## Final Answer for 3: The projectile has zero horizontal acceleration and a vertical acceleration of \(-g\). These mechanics apply due to the independence of horizontal and vertical motions in projectile motion. --- ## 4. **Plot qualitatively the trajectory for both cases (free flight and accelerating base)** ### **Step 1: Trajectory for Free Flight** - **Free Flight**: The projectile follows a parabolic trajectory due to gravity acting alone. ### **Step 2: Trajectory for Accelerating Base** - **Accelerating Base**: The trajectory will appear more horizontal than the free flight due to the initial horizontal velocity from the platform and the additional horizontal component from the projectile's launch. ### **Qualitative Plot** ```plaintext | * | * | * | * | * | * | * |______________________ ``` - The left curve represents the **free flight** trajectory. - The right curve shows the **trajectory from the accelerating base**, with a steeper drop due to the initial horizontal velocity component. **Explanation:** The projectile's path changes based on its initial conditions, including the motion of the platform. The horizontal acceleration of the base affects the apparent trajectory observed from a stationary point. --- ## Final Answer for 4: The trajectories differ qualitatively, where the free flight follows a standard parabolic path, while the accelerating base creates a flatter trajectory due to the initial horizontal velocity imparted by the platform.