At serve, a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear the 0.90-m-high-net about 15.0 m from the server if the ball is "launched" from a height of 2.50 m? Where will the ball land if it just clears the net (and will it be "good "in the sense that it lands within 7.0m of the net)? How long will it be in the air? See Fig. 3-54

To solve this problem, we need to use the equations of motion for a projectile. Let's break down the problem into three parts: finding the minimum speed required to clear the net, determining where the ball will land, and calculating the time the ball will be in the air.<br /><br />### Part 1: Minimum Speed Required to Clear the Net<br /><br />We need to find the initial speed \( v_0 \) such that the ball clears the net by at least 0.15 m (since the net is 0.90 m high and the ball is launched from 2.50 m high, the clearance needed is 0.60 m).<br /><br />The vertical displacement \( \Delta y \) is given by:<br />\[ \Delta y = y_f - y_i = 0.90 \, \text{m} - 2.50 \, \text{m} = -1.60 \, \text{m} \]<br /><br />The horizontal displacement \( \Delta x \) is given by:<br />\[ \Delta x = v_0 t \]<br /><br />The time \( t \) can be found using the vertical motion equation:<br />\[ \Delta y = v_{0y} t + \frac{1}{2} a t^2 \]<br />where \( v_{0y} = v_0 \sin(\theta) \) and \( a = -g \).<br /><br />Since we are dealing with a horizontal launch, \( \theta = 0 \), so \( v_{0y} = 0 \). Thus, the equation simplifies to:<br />\[ \Delta y = -\frac{1}{2} g t^2 \]<br /><br />Solving for \( t \):<br />\[ t = \sqrt{\frac{2 \Delta y}{-g}} \]<br />\[ t = \sqrt{\frac{2 \times 1.60 \, \text{m}}{9.8 \, \text{m/s}^2}} \]<br />\[ t = \sqrt{\frac{3.20 \, \text{m}}{9.8 \, \text{m/s}^2}} \]<br />\[ t = \sqrt{0.3265 \, \text{s}^2} \]<br />\[ t \approx 0.571 \, \text{s} \]<br /><br />Now, using this time in the horizontal motion equation:<br />\[ \Delta x = v_0 t \]<br />\[ 15.0 \, \text{m} = v_0 \times 0.571 \, \text{s} \]<br />\[ v_0 = \frac{15.0 \, \text{m}}{0.571 \, \text{s}} \]<br />\[ v_0 \approx 26.3 \, \text{m/s} \]<br /><br />### Part 2: Where Will the Ball Land?<br /><br />Using the horizontal motion equation:<br />\[ \Delta x = v_0 t \]<br />\[ \Delta x = 26.3 \, \text{m/s} \times 0.571 \, \text{s} \]<br />\[ \Delta x \approx 14.9 \, \text{m} \]<br /><br />Since 14.9 m is within 7.0 m of the net, the ball will land "good."<br /><br />### Part 3: Time in the Air<br /><br />The total time of flight is twice the time to reach the net height:<br />\[ t_{\text{total}} = 2 \times 0.571 \, \text{s} \]<br />\[ t_{\text}} \approx 1.14 \, \text{s} \]<br /><br />### Summary<br /><br />- **Minimum Speed Required:** \( 26.3 \, \text{m/s} \)<br />- **Landing Location:** Approximately 14.9 m from the net (within 7.0 m)<br />- **Total Time in the Air:** Approximately 1.14 s