Let's solve each question step by step. --- ## **Question 1** **Given:** Lighthouse height above water = 49 ft - Cliff extends from base = 19 ft (horizontal distance from base shore) - Sailor's eye above water = 14 ft - Angle above horizontal to top of lighthouse = \(30^\circ\) - Find: Distance from ship to the rocks (horizontal distance from ship to base of cliff). ### **Step 1: Find the vertical distance between sailor's eye and top of lighthouse** \[ \text{Vertical difference} = 49\,\text{ft} - 14\,\text{ft} = 35\,\text{ft} \] ### **Step 2: Use trigonometry** Let \( x \) = horizontal distance from sailor to base of lighthouse. From the right triangle: \[ \tan(30^\circ) = \frac{\text{vertical difference}}{x} \] \[ \tan(30^\circ) = \frac{35}{x} \] \[ x = \frac{35}{\tan(30^\circ)} \] \[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx .577 \] \[ x = \frac{35}{.577} \approx 60.66\, \text{ft} \] ### **Step 3: Add the distance from base of cliff to rocks** Total distance from ship to rocks: \[ \text{Total distance} = x + 19 = 60.66 + 19 = 79.66\, \text{ft} \] **Final Answer:** \[ \boxed{79.7\, \text{ft}} \] --- ## **Question 2** **Given:** Find the magnitude and direction of the resultant force. \[ \begin{align*} F_1 & = 40\, \text{N} \, (\text{up}) \\ F_2 & = 50\, \text{N} \, (60^\circ \text{ right of up}) \\ F_3 & = 30\, \text{N} \, (30^\circ \text{ right of down}) \\ F_4 & = 50\, \text{N} \, (down-left, 60^\circ \text{ left of down}) \\ \end{align*} \] ### **Step 1: Resolve all forces into components** Let up = positive \(y\), right = positive \(x\): #### \(F_1\) (up) \[ F_{1x} = , \quad F_{1y} = 40 \] #### \(F_2\) (60° to the right of up) \[ F_{2x} = 50 \sin(60^\circ) = 50 \times .866 = 43.3 \] \[ F_{2y} = 50 \cos(60^\circ) = 50 \times .5 = 25 \] #### \(F_3\) (30° to the right of down) \[ F_{3x} = 30 \sin(30^\circ) = 30 \times .5 = 15 \] \[ F_{3y} = -30 \cos(30^\circ) = -30 \times .866 = -25.98 \] #### \(F_4\) (60° to the left of down) \[ F_{4x} = -50 \sin(60^\circ) = -50 \times .866 = -43.3 \] \[ F_{4y} = -50 \cos(60^\circ) = -50 \times .5 = -25 \] ### **Step 2: Sum up all \(x\) and \(y\) components** \[ F_x = + 43.3 + 15 - 43.3 = 15 \] \[ F_y = 40 + 25 - 25.98 - 25 = 40 + 25 - 25.98 - 25 = 14.02 \] ### **Step 3: Magnitude and direction** \[ F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{15^2 + 14.02^2} = \sqrt{225 + 196.56} = \sqrt{421.56} = 20.53\, \text{N} \] \[ \theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{14.02}{15}\right) = \tan^{-1}(.9347) \approx 43.4^\circ \text{ above the } x\text{-axis} \] **Final Answer:** \[ \boxed{20.5\, \text{N}, \quad 43.4^\circ \text{ above the } x\text{-axis}} \] --- ## **Question 3** ### **(a) Time to hit the floor** Given: - Height, \( h = 1.00\,\text{m} \) - Initial vertical velocity, \( v_{y} = \) - \( g = 9.8\,\text{m/s}^2 \) \[ h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.00}{9.8}} = \sqrt{.204} = .452\,\text{s} \] ### **(b) How far from the bench will the ball land?** Horizontal speed \( v_x = 1.25\,\text{m/s} \): \[ d = v_x \times t = 1.25 \times .452 = .565\,\text{m} \] **Final Answers:** - (a) \( \boxed{.45\,\text{s}} \) - (b) \( \boxed{.57\,\text{m}} \) --- ## **Question 4** ### **Step 1: Distance traveled during reaction time** \[ d_1 = v \times t_{reaction} = 20 \times .5 = 10\,\text{m} \] ### **Step 2: Braking distance** Maximum deceleration \( a = -6.\,\text{m/s}^2 \), initial speed \(v_ = 20\,\text{m/s}\), final speed \(v = \): \[ v^2 = v_^2 + 2ad \] \[ = (20)^2 + 2(-6)d_2 \] \[ = 400 - 12d_2 \] \[ 12d_2 = 400 \implies d_2 = \frac{400}{12} \approx 33.33\,\text{m} \] ### **Step 3: Total stopping distance** \[ d_{total} = d_1 + d_2 = 10 + 33.33 = 43.33\,\text{m} \] Since the obstacle is 50 m ahead, and stopping distance is 43.33 m: **Final Answer:** \[ \boxed{\text{Yes, she can stop in time.}} \]