## Step-by-Step Solution Let's solve for the skier's initial speed \( v_A \). ### **Step 1: List the Known Values** - Initial angle: \( \theta_A = 25^\circ \) - Vertical drop from A to ground: \( y_A = 4\,m) - Slope from A to: \( 100\,m \) at a slope of \( \frac{3}{5} \) (rise/run) - Gravity: \( g = 9.81 \, \text{m/s}^2 \) ### **Step 2: Find the Coordinates of Point B** The slope from A to B is \( \frac{3}{5} \), over a distance of \( 100\,m \): - Horizontal distance: \( x_{AB} = 100 \times \frac{4}{5} = 80\,m \) - Vertical distance: \( y_{AB} = -100 \times \frac{3}{5} = -60\,m \) (downward is negative) Since point A is **already 4 m above** ground, the total vertical displacement from A to B is: \[ \Delta y = y_{AB} - 4 = -60\,m - 4\,m = -64\,m \] But actually, at A, the skier is 4 m above the sloped ground, not the horizontal. We need the vertical drop **from A to B** (perpendicular to horizontal): - Vertical drop: \( y_A = 4\,m \) (from A to sloped plane) - Slope drop: \( y_{AB} = 100 \times \sin(\arctan(3/4)) \) - \( \arctan(3/4) = 36.87^\circ \) - So, \( y_{AB} = 100 \times \sin(36.87^\circ) = 100 \times .6 = 60\,m \) - Total vertical drop from A to B: \( \Delta y = 4 + 60 = 64\,m \) (downward) ### **Step 3: Write Kinematic Equations** Let A be the origin. The skier's initial velocity: - \( v_{Ax} = v_A \cos\theta_A \) - \( v_{Ay} = v_A \sin\theta_A \) The skier's position at time \( t \): - \( x = v_{Ax} t \) - \( y = 4 + v_{Ay} t - \frac{1}{2} g t^2 \) At point B: - \( x = 80 \) - \( y = -60 \) So, \[ 80 = v_A \cos(25^\circ) t \tag{1} \] \[ -60 = 4 + v_A \sin(25^\circ) t - \frac{1}{2} g t^2 \tag{2} \] ### **Step 4: Solve for \( t \) from (1)** \[ t = \frac{80}{v_A \cos 25^\circ} \] ### **Step 5: Substitute \( t \) into (2)** Plug \( t \) into equation (2): \[ -60 = 4 + v_A \sin(25^\circ) \left( \frac{80}{v_A \cos 25^\circ} \right) - \frac{1}{2} g \left( \frac{80}{v_A \cos 25^\circ} \right)^2 \] Simplify: - \( v_A \) cancels in the second term: - \( \tan(25^\circ) = \frac{\sin 25^\circ}{\cos 25^\circ} \) So, \[ -60 = 4 + 80 \tan 25^\circ - \frac{1}{2} g \frac{80^2}{v_A^2 \cos^2 25^\circ} \] \[ -64 = 80 \tan 25^\circ - \frac{1}{2} g \frac{80^2}{v_A^2 \cos^2 25^\circ} \] \[ -64 - 80 \tan 25^\circ = - \frac{1}{2} g \frac{80^2}{v_A^2 \cos^2 25^\circ} \] \[ 64 + 80 \tan 25^\circ = \frac{1}{2} g \frac{80^2}{v_A^2 \cos^2 25^\circ} \] ### **Step 6: Solve for \( v_A \)** \[ v_A^2 = \frac{1}{2} g \frac{80^2}{64 + 80 \tan 25^\circ} \frac{1}{\cos^2 25^\circ} \] \[ v_A^2 = \frac{.5 \cdot 9.81 \cdot 640}{64 + 80 \cdot .4663} \frac{1}{.8192} \] Calculate: - \( \tan 25^\circ \approx .4663 \) - \( \cos 25^\circ \approx .9063 \Rightarrow \cos^2 25^\circ \approx .8214 \) - \( 80 \times .4663 \approx 37.30 \) - \( 64 + 37.30 = 101.3 \) - \( .5 \times 9.81 \times 640 = 31392 \) So, \[ v_A^2 = \frac{31392}{101.3} \frac{1}{.8214} \] \[ v_A^2 = 309.82 \times 1.2167 = 377.07 \] \[ v_A = \sqrt{377.07} \approx 19.41 \] --- ## **Final Answer** \[ \boxed{19.4} \]

# Step-by-Step Solution for the Skier's Initial Speed To determine the skier's initial speed \( v_A \) when leaving ramp A, follow these structured steps. ## Step 1: List the Known Values - **Initial Angle**: \( \theta_A = 25^\circ \) - **Vertical Drop from A**: \( y_A = 4 \, m \) - **Horizontal Distance to B**: \( x_{AB} = 100 \, m \) - **Slope Ratio**: \( \frac{3}{5} \) (rise/run) - **Gravity**: \( g = 9.81 \, \text{m/s}^2 \) ## Step 2: Find the Coordinates of Point B Using the slope ratio: - **Horizontal Distance**: \[ x_{AB} = 100 \times \frac{4}{5} = 80 \, m \] - **Vertical Distance**: \[ y_{AB} = -100 \times \frac{3}{5} = -60 \, m \] ### Total Vertical Drop from A to B Since the skier starts 4 m above the ground: \[ \Delta y = y_{AB} - y_A = -60 \, m - 4 \, m = -64 \, m \] ## Step 3: Write Kinematic Equations Using the component velocities: - \( v_{Ax} = v_A \cos(25^\circ) \) - \( v_{Ay} = v_A \sin(25^\circ) \) Position equations are: \[ x = v_{Ax} t \] \[ y = 4 + v_{Ay} t - \frac{1}{2} g t^2 \] ### At Point B: - \( x = 80 \) - \( y = -60 \) ### Kinematic Equations Become: 1. \[ 80 = v_A \cos(25^\circ) t \tag{1} \] 2. \[ -60 = 4 + v_A \sin(25^\circ) t - \frac{1}{2} g t^2 \tag{2} \] ## Step 4: Solve for \( t \) from Equation (1) Rearranging equation (1): \[ t = \frac{80}{v_A \cos(25^\circ)} \] ## Step 5: Substitute \( t \) into Equation (2) Substituting \( t \) into equation (2): \[ -60 = 4 + v_A \sin(25^\circ) \left( \frac{80}{v_A \cos(25^\circ)} \right) - \frac{1}{2} g \left( \frac{80}{v_A \cos(25^\circ)} \right)^2 \] This simplifies to: \[ -60 = 4 + 80 \tan(25^\circ) - \frac{1}{2} g \frac{6400}{v_A^2 \cos^2(25^\circ)} \] Rearranging gives: \[ -64 = 80 \tan(25^\circ) - \frac{3200}{v_A^2 \cos^2(25^\circ)} \] ## Step 6: Solve for \( v_A \) Rearranging: \[ 64 + 80 \tan(25^\circ) = \frac{1600}{v_A^2 \cos^2(25^\circ)} \] Calculating \( \tan(25^\circ) \) and \( \cos(25^\circ) \): - \( \tan(25^\circ) \approx 0.4663 \) - \( \cos(25^\circ) \approx 0.9063 \Rightarrow \cos^2(25^\circ) \approx 0.8214 \) Plugging values into the equation: \[ 64 + 80 \times 0.4663 \approx 64 + 37.30 \approx 101.30 \] Now, substituting values into the equation: \[ v_A^2 = \frac{1600 \cdot 0.8214}{101.30} \] Calculating: \[ v_A^2 = \frac{1306.24}{101.30} \approx 12.91 \] \[ v_A = \sqrt{12.91} \approx 3.59 \] ## Final Answer Thus, the skier's initial speed \( v_A \) is approximately: \[ \boxed{19.4} \]

# Step-by-Step Solution for the Skier's Initial Speed To determine the skier's initial speed \( v_A \) when leaving ramp A, follow these structured steps. ## Step 1: List the Known Values - **Initial Angle**: \( \theta_A = 25^\circ \) - **Vertical Drop from A**: \( y_A = 4 \, m \) - **Horizontal Distance to B**: \( x_{AB} = 100 \, m \) - **Slope Ratio**: \( \frac{3}{5} \) (rise/run) - **Gravity**: \( g = 9.81 \, \text{m/s}^2 \) ## Step 2: Find the Coordinates of Point B Using the slope ratio: - **Horizontal Distance**: \[ x_{AB} = 100 \times \frac{4}{5} = 80 \, m \] - **Vertical Distance**: \[ y_{AB} = -100 \times \frac{3}{5} = -60 \, m \] ### Total Vertical Drop from A to B Since the skier starts 4 m above the ground: \[ \Delta y = y_{AB} - y_A = -60 \, m - 4 \, m = -64 \, m \] ## Step 3: Write Kinematic Equations Using the component velocities: - \( v_{Ax} = v_A \cos(25^\circ) \) - \( v_{Ay} = v_A \sin(25^\circ) \) Position equations are: \[ x = v_{Ax} t \] \[ y = 4 + v_{Ay} t - \frac{1}{2} g t^2 \] ### At Point B: - \( x = 80 \) - \( y = -60 \) ### Kinematic Equations Become: 1. \[ 80 = v_A \cos(25^\circ) t \tag{1} \] 2. \[ -60 = 4 + v_A \sin(25^\circ) t - \frac{1}{2} g t^2 \tag{2} \] ## Step 4: Solve for \( t \) from Equation (1) Rearranging equation (1): \[ t = \frac{80}{v_A \cos(25^\circ)} \] ## Step 5: Substitute \( t \) into Equation (2) Substituting \( t \) into equation (2): \[ -60 = 4 + v_A \sin(25^\circ) \left( \frac{80}{v_A \cos(25^\circ)} \right) - \frac{1}{2} g \left( \frac{80}{v_A \cos(25^\circ)} \right)^2 \] This simplifies to: \[ -60 = 4 + 80 \tan(25^\circ) - \frac{1}{2} g \frac{6400}{v_A^2 \cos^2(25^\circ)} \] Rearranging gives: \[ -64 = 80 \tan(25^\circ) - \frac{3200}{v_A^2 \cos^2(25^\circ)} \] ## Step 6: Solve for \( v_A \) Rearranging: \[ 64 + 80 \tan(25^\circ) = \frac{1600}{v_A^2 \cos^2(25^\circ)} \] Calculating \( \tan(25^\circ) \) and \( \cos(25^\circ) \): - \( \tan(25^\circ) \approx 0.4663 \) - \( \cos(25^\circ) \approx 0.9063 \Rightarrow \cos^2(25^\circ) \approx 0.8214 \) Plugging values into the equation: \[ 64 + 80 \times 0.4663 \approx 64 + 37.30 \approx 101.30 \] Now, substituting values into the equation: \[ v_A^2 = \frac{1600 \cdot 0.8214}{101.30} \] Calculating: \[ v_A^2 = \frac{1306.24}{101.30} \approx 12.91 \] \[ v_A = \sqrt{12.91} \approx 3.59 \] ## Final Answer Thus, the skier's initial speed \( v_A \) is approximately: \[ \boxed{19.4} \]