As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. Draw the figure in the space provided below and, showing your work, answer the following questions:
(a)Determine the initial velocity of the ball in the x- and y-direction.
(b) What is the acceleration of the ball in the x- and y-direction?
(c) How much time does the ball spend in air?
(d) How far above the court is the tennis ball when it leaves the racket?
(e) What is the x-component of the ball’s velocity when the ball hits the court?
(f) What is the y-component of the ball’s velocity when the ball hits the court?
(g) What is the velocity when the ball hits the court?
The correct answer and explanation is :
Given Information:
- The initial horizontal velocity of the ball, $v_x = 28.0 \, \text{m/s}$.
- The horizontal distance the ball travels, $x = 19.6 \, \text{m}$.
We need to find the velocity components and other parameters of the ball’s motion.
(a) Determine the initial velocity of the ball in the x- and y-direction.
- Initial velocity in the x-direction, $v_{x0}$:
- The horizontal velocity remains constant because there is no horizontal acceleration (neglecting air resistance).
- Hence, $v_{x0} = 28.0 \, \text{m/s}$.
- Initial velocity in the y-direction, $v_{y0}$:
- Since the ball is struck horizontally, there is no initial vertical velocity.
- Therefore, $v_{y0} = 0 \, \text{m/s}$.
(b) What is the acceleration of the ball in the x- and y-direction?
- Acceleration in the x-direction, $a_x$:
- In the absence of air resistance, the horizontal velocity does not change.
- Hence, $a_x = 0 \, \text{m/s}^2$.
- Acceleration in the y-direction, $a_y$:
- The only force acting vertically on the ball is gravity, causing a downward acceleration.
- Therefore, $a_y = -9.8 \, \text{m/s}^2$.
(c) How much time does the ball spend in the air?
The time of flight can be determined by considering the vertical motion. Since the ball was struck horizontally, its initial vertical velocity is zero.
Using the equation for vertical displacement:
$$
y = v_{y0} t + \frac{1}{2} a_y t^2
$$
Since $y = 0$ (the ball returns to the court), we get:
$$
0 = 0 + \frac{1}{2} (-9.8) t^2
$$
Simplifying:
$$
t^2 = \frac{2y}{g}
$$
Since the ball starts at some height $y_0$ above the court (which we will later find), we need a different approach to solve for time. First we’ll solve the equation numerically