Drop shot 2

A "drop shot" in tennis is a shot that barely passes the net and touches the other side of the court close to the net. This makes it difficult for the opponent to reach the ball if the opponent starts running for the ball from the baseline. A drop shot also has a backspin that makes the ball bounce off the ground in a funny way so that the opponent will have an even harder time. In this problem, we will look at this second part: the backspin. (See this for the first part without a backspin.)

You've hit a ball with a really strong backspin. The ball touches the ground $0.9\text{ m}$ from the net on the opponent's side with a velocity of $6.9\text{ m/s}$ at an angle of $54^{\circ}$ with the ground. It still spins backward when it bounces up. The coefficient of friction is $\mu=0.50$ and the coefficient of restitution is $e=0.80$ .

Where will the ball touch the ground again?

Details and Assumptions:

• in a backspin, the bottom of the ball moves forward and the top of the ball moves backward, relative to the velocity of the center of the ball.
• Neglect air resistance.
• Use $g=10\text{ m/s}^2$ .
• The plane containing the trajectory of the ball is perpendicular to the net.
• In a collision between a massive, horizontal, flat surface and a ball, the coefficient of restitution is defined as $e=\left|\dfrac{v_y'}{v_y}\right|,$ where $v_y$ and $v_y'$ are the vertical components of the velocity of the ball, before and after the collision, respectively.