Let's solve part (a) step by step: ### **Problem Recap:** - Initial speed of cannonball, \( v_0 = 50 \text{ m/s} \) - Angle of launch, \( \theta = 30^\circ \) We need to find the **x** (horizontal) and **y** (vertical) components of the initial velocity. --- ### **Step 1: Draw the Right Triangle** Imagine the initial velocity vector (\( v_0 \)) as the hypotenuse of a right triangle: - The **horizontal leg** (adjacent to the angle): \( v_{0x} \) - The **vertical leg** (opposite the angle): \( v_{0y} \) - The **angle** with the horizontal is \( 30^\circ \) ``` | |\ | \ v0y | \ v0 = 50 m/s | \ |____\ v0x (angle = 30°) ``` --- ### **Step 2: Write the Equations for Components** The components are given by: - \( v_{0x} = v_0 \cos\theta \) - \( v_{0y} = v_0 \sin\theta \) --- ### **Step 3: Substitute Values** Given \( v_0 = 50 \text{ m/s} \), \( \theta = 30^\circ \): - \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \) - \( \sin(30^\circ) = \frac{1}{2} = 0.5 \) So: - \( v_{0x} = 50 \times 0.866 = 43.3 \text{ m/s} \) - \( v_{0y} = 50 \times 0.5 = 25.0 \text{ m/s} \) --- ### **Final Answer** **The x and y components of the initial velocity are:** \[ \boxed{ \begin{aligned} v_{0x} &= 43.3 \ \text{m/s} \\ v_{0y} &= 25.0 \ \text{m/s} \end{aligned} } \] Let me know if you'd like a solution to the next part!

Certainly! Let's proceed step-by-step to solve the entire problem, including parts (a), (b), and (c), with detailed calculations and explanations. --- ## Given Data: - Initial velocity, \( v_0 = 50\, \text{m/s} \) - Launch angle, \( \theta = 30^\circ \) - Height of the fort's top above water, \( h = 10\, \text{meters} \) - Acceleration due to gravity, \( g = 9.8\, \text{m/s}^2 \) --- ## Part (a): Find the x and y components of the initial velocity ### Step 1: Draw the velocity triangle (conceptual, as previously explained). ### Step 2: Compute components: \[ v_{0x} = v_0 \cos \theta = 50 \times \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} \approx 50 \times 0.866 = \boxed{43.3\, \text{m/s}} \] \[ v_{0y} = v_0 \sin \theta = 50 \times \frac{1}{2} = \boxed{25.0\, \text{m/s}} \] --- ## Part (b): Find the time of flight until the projectile hits the water level (water is at height 0), considering the projectile is launched from 10 meters above water. ### Step 1: Set up the vertical motion equation Vertical position as a function of time: \[ y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2 \] Where: - \( y_0 = 10\, \text{m} \) (initial height above water) - \( y(t) = 0 \) (water level) - \( v_{0y} = 25\, \text{m/s} \) ### Step 2: Write the quadratic equation: \[ 0 = 10 + 25 t - \frac{1}{2} \times 9.8 \times t^2 \] \[ 0 = 10 + 25 t - 4.9 t^2 \] Rearranged as: \[ 4.9 t^2 - 25 t - 10 = 0 \] ### Step 3: Solve the quadratic: \[ t = \frac{25 \pm \sqrt{(-25)^2 - 4 \times 4.9 \times (-10)}}{2 \times 4.9} \] Calculate discriminant: \[ \Delta = 625 + 196 = 821 \] Calculate square root: \[ \sqrt{821} \approx 28.66 \] Now, find the roots: \[ t = \frac{25 \pm 28.66}{9.8} \] Two solutions: - \( t_1 = \frac{25 + 28.66}{9.8} = \frac{53.66}{9.8} \approx 5.48\, \text{s} \) - \( t_2 = \frac{25 - 28.66}{9.8} = \frac{-3.66}{9.8} \approx -0.37\, \text{s} \) (discard negative time) **Thus, the projectile hits the water after approximately:** \[ \boxed{t \approx 5.48\, \text{s}} \] --- ## Part (c): Find the horizontal distance (range) when the projectile hits the water ### Step 1: Horizontal motion Horizontal velocity remains constant: \[ v_{0x} = 43.3\, \text{m/s} \] Horizontal distance traveled: \[ x = v_{0x} \times t = 43.3 \times 5.48 \approx 237.5\, \text{meters} \] --- ## **Final Summary:** | Part | Calculation | Result | |---|---|---| | **(a)** | Components of initial velocity: | \( v_{0x} \approx 43.3\, \text{m/s} \) <br> \( v_{0y} \approx 25.0\, \text{m/s} \) | | **(b)** | Time to hit water: | \( t \approx 5.48\, \text{s} \) | | **(c)** | Horizontal distance traveled: | \( x \approx 237.5\, \text{meters} \) | --- **Note:** This analysis assumes no air resistance and that the projectile lands exactly at water level after 5.48 seconds.

Let's analyze the problem step-by-step, covering all parts systematically with detailed explanations and calculations. --- ### **Given Data:** - Initial speed, \( v_0 = 50\, \text{m/s} \) - Launch angle, \( \theta = 30^\circ \) - Height of the fort's top above water, \( y_0 = 10\, \text{m} \) - Gravitational acceleration, \( g = 9.8\, \text{m/s}^2 \) --- ## **Part (a): Find the x and y components of the initial velocity** ### **Step 1: Draw the velocity triangle** The initial velocity vector \( v_0 \) can be decomposed into: - Horizontal component \( v_{0x} \) - Vertical component \( v_{0y} \) Using the right triangle: ``` |\ | \ | \ v0 = 50 m/s | \ |____\ v0x ``` ### **Step 2: Calculate components** \[ v_{0x} = v_0 \cos \theta = 50 \times \cos 30^\circ \] \[ v_{0y} = v_0 \sin \theta = 50 \times \sin 30^\circ \] Recall: \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866,\quad \sin 30^\circ = 0.5 \] Thus: \[ v_{0x} = 50 \times 0.866 = \boxed{43.3\, \text{m/s}} \] \[ v_{0y} = 50 \times 0.5 = \boxed{25.0\, \text{m/s}} \] --- ## **Part (b): Find the time \( t \) when the projectile hits the water** ### **Step 1: Vertical motion equation** Vertical position at time \( t \): \[ y(t) = y_0 + v_{0y} t - \frac{1}{2} g t^2 \] Given: \[ y(t) = 0 \quad \text{(water level)},\quad y_0=10\, \text{m} \] \[ 0 = 10 + 25 t - 4.9 t^2 \] ### **Step 2: Solve the quadratic** Rearranged: \[ 4.9 t^2 - 25 t - 10 = 0 \] Discriminant: \[ \Delta = (-25)^2 - 4 \times 4.9 \times (-10) = 625 + 196 = 821 \] Square root: \[ \sqrt{821} \approx 28.66 \] Solutions: \[ t = \frac{25 \pm 28.66}{2 \times 4.9} = \frac{25 \pm 28.66}{9.8} \] - Positive root: \[ t = \frac{25 + 28.66}{9.8} = \frac{53.66}{9.8} \approx \boxed{5.48\, \text{s}} \] - Negative root (discarded, as time can't be negative). **Result:** The projectile hits water after approximately **5.48 seconds**. --- ## **Part (c): Find the horizontal distance traveled when it hits water** ### **Step 1: Horizontal motion** Horizontal velocity \( v_{0x} \) remains constant: \[ v_{0x} = 43.3\, \text{m/s} \] ### **Step 2: Calculate the range** \[ x = v_{0x} \times t = 43.3 \times 5.48 \approx 237.5\, \text{meters} \] --- ## **Final answers summary:** | Part | Result | |---|---| | **(a)** | \( v_{0x} \approx 43.3\, \text{m/s} \), \( v_{0y} \approx 25.0\, \text{m/s} \) | | **(b)** | Time of flight until water impact \( \approx 5.48\, \text{s} \) | | **(c)** | Horizontal distance traveled \( \approx 237.5\, \text{meters} \) | --- ### **Additional note:** **Final velocity just before impact:** - Horizontal component remains \( v_{x} = 43.3\, \text{m/s} \) (constant). - Vertical component at impact: \[ v_{y} = v_{0y} - g t = 25 - 9.8 \times 5.48 \approx 25 - 53.7 = -28.7\, \text{m/s} \] (The negative sign indicates downward direction.) **Magnitude of velocity just before impact:** \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(43.3)^2 + (-28.7)^2} \approx \sqrt{1875 + 823} \approx \sqrt{2698} \approx 52\, \text{m/s} \] **Direction (angle below horizontal):** \[ \theta_{impact} = \arctan \left( \frac{|v_y|}{v_x} \right) = \arctan \left( \frac{28.7}{43.3} \right) \approx \arctan(0.66) \approx 33^\circ \] --- **Let me know if you'd like further clarification or additional parts!**