A tennis player standing 17.7 m from the net hits the ball at 5.00 degrees above the horizontal. To clear the net, the ball must rise at least 0.275 m . If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?

Alright, I need to determine the initial speed of the tennis ball when it was hit by the player. The problem provides the following information: - The distance from the net is 17.7 meters. - The ball is hit at an angle of 5.00 degrees above the horizontal. - The ball must rise at least 0.275 meters to clear the net. - The ball just clears the net at the apex of its trajectory. First, I'll visualize the scenario. The tennis player is standing 17.7 meters away from the net, and the ball is hit at a 5-degree angle. The ball needs to reach a height of at least 0.275 meters to clear the net. Since the ball just clears the net at the apex of its trajectory, this means that at the highest point of its flight, the vertical component of its velocity is zero, and it has reached the required height. To solve this problem, I'll break it down into two main parts: determining the vertical and horizontal components of the initial velocity and then using the kinematic equations to find the initial speed. **Step 1: Define the Variables** Let: - \( v_0 \) = initial speed of the ball (what we need to find) - \( \theta = 5.00^\circ \) = angle of projection above the horizontal - \( d = 17.7 \) meters = horizontal distance to the net - \( h = 0.275 \) meters = minimum height to clear the net - \( g = 9.81 \) m/s² = acceleration due to gravity **Step 2: Resolve the Initial Velocity into Components** The initial velocity can be resolved into horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)) components: \[ v_{0x} = v_0 \cos(\theta) \] \[ v_{0y} = v_0 \sin(\theta) \] **Step 3: Determine the Time of Flight to the Net** Since the ball just clears the net at the apex, the time it takes to reach the net (\( t \)) can be found using the horizontal motion: \[ d = v_{0x} \times t \] \[ t = \frac{d}{v_{0x}} = \frac{17.7}{v_0 \cos(5.00^\circ)} \] **Step 4: Determine the Vertical Velocity at the Apex** At the apex, the vertical component of the velocity is zero. Using the kinematic equation: \[ v_{y} = v_{0y} - g \times t \] \[ 0 = v_0 \sin(5.00^\circ) - g \times t \] \[ t = \frac{v_0 \sin(5.00^\circ)}{g} \] **Step 5: Equate the Two Expressions for Time** Set the two expressions for \( t \) equal to each other: \[ \frac{17.7}{v_0 \cos(5.00^\circ)} = \frac{v_0 \sin(5.00^\circ)}{g} \] **Step 6: Solve for \( v_0 \)** Rearranging the equation: \[ 17.7 \times g = v_0^2 \times \sin(5.00^\circ) \times \cos(5.00^\circ) \] \[ v_0^2 = \frac{17.7 \times g}{\sin(5.00^\circ) \times \cos(5.00^\circ)} \] \[ v_0 = \sqrt{\frac{17.7 \times 9.81}{\sin(5.00^\circ) \times \cos(5.00^\circ)}} \] **Step 7: Calculate the Numerical Value** First, calculate \( \sin(5.00^\circ) \) and \( \cos(5.00^\circ) \): \[ \sin(5.00^\circ) \approx 0.0872 \] \[ \cos(5.00^\circ) \approx 0.9962 \] Now, plug these values into the equation: \[ v_0 = \sqrt{\frac{17.7 \times 9.81}{0.0872 \times 0.9962}} \] \[ v_0 = \sqrt{\frac{173.537}{0.0869}} \] \[ v_0 = \sqrt{1999.99} \] \[ v_0 \approx 44.7 \text{ m/s} \] **Step 8: Verify the Calculation** To ensure accuracy, let's double-check the calculations: \[ \sin(5.00^\circ) \approx 0.0872 \] \[ \cos(5.00^\circ) \approx 0.9962 \] \[ \sin(5.00^\circ) \times \cos(5.00^\circ) \approx 0.0872 \times 0.9962 \approx 0.0869 \] \[ 17.7 \times 9.81 = 173.537 \] \[ \frac{173.537}{0.0869} \approx 1999.99 \] \[ \sqrt{1999.99} \approx 44.7 \text{ m/s} \] The calculations seem consistent and accurate. **Conclusion** The tennis ball was moving at approximately **44.7 meters per second** when it left the racket.