A tennis ball is struck and departs from the racket horizontally with a speed of 28.6 m/s. The ball hits the court at a horizontal distance of 18.4 m from the racket. How far above the court is the tennis ball when it leaves the racket?
To solve this problem, we can use the kinematic equation for horizontal motion:
Horizontal distance (x) = Initial horizontal velocity (v_x) × Time (t)
Given:
Initial horizontal velocity (v_x) = 28.6 m/s
Horizontal distance (x) = 18.4 m
We need to find the time (t) it takes for the ball to reach the horizontal distance of 18.4 m.
Using the formula, we can rearrange it to solve for time (t):
t = x / v_x
Plugging in the values:
t = 18.4 m / 28.6 m/s
t ≈ 0.643 s
The time it takes for the ball to reach the horizontal distance of 18.4 m is approximately 0.643 s.
Next, we need to find the vertical distance (y) traveled by the ball when it leaves the racket.
We can use the vertical motion equations of a freely falling object:
y = v_y × t + 0.5 × g × t^2
Since the object is struck horizontally, the initial vertical velocity (v_y) is 0 m/s.
Also, the acceleration due to gravity (g) is approximately 9.8 m/s^2.
Plugging in the values and solving for y:
y = 0 × 0.643 s + 0.5 × 9.8 m/s^2 × (0.643 s)^2
y ≈ 0.994 m
The tennis ball is approximately 0.994 m above the court when it leaves the racket.
To determine how far above the court the tennis ball is when it leaves the racket, we can use the equations of motion in projectile motion.
Let's consider the horizontal and vertical components of the motion separately.
First, let's analyze the horizontal component. The horizontal speed of the ball remains constant throughout its flight. We are given that the horizontal distance traveled by the ball is 18.4 m. The formula for horizontal distance (d) is given by:
d = v_x * t
where v_x is the horizontal component of the velocity and t is the time of flight. Since there is no horizontal acceleration (assuming neglecting air resistance), we can say that v_x is constant. Therefore, we can rearrange the formula to solve for time:
t = d / v_x
where d = 18.4 m and v_x is the horizontal component of the initial velocity (which is 28.6 m/s). Substituting the given values:
t = 18.4 m / 28.6 m/s ≈ 0.643 seconds
Now, let's move on to the vertical component of motion. We need to determine the initial vertical velocity (v_y) of the ball when it leaves the racket. Since the ball is struck horizontally, there is no initial vertical velocity (v_y = 0 m/s).
Next, we can use the equation of motion to calculate the time it takes for the ball to reach the highest point of its trajectory (when the vertical velocity becomes zero). The equation is:
v_y = v_y0 + g * t
where v_y is the vertical component of the velocity, v_y0 is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight (which we calculated earlier as 0.643 seconds).
0 = 0 + (-9.8 m/s²) * t
Simplifying the equation:
t = 0
Since the equation yields t = 0, it means the ball reaches its highest point instantaneously. This indicates that the time taken for the ball to reach its highest point is the same as the time taken for it to reach the ground.
Now, we can use the equation to solve for the vertical displacement of the ball:
d_y = v_y0 * t + (1/2) * g * t²
Since v_y0 is 0 m/s, the equation simplifies to:
d_y = (1/2) * g * t²
Substituting the values:
d_y = (1/2) * (-9.8 m/s²) * (0.643 s)² ≈ -2.83 m
The negative sign indicates that the ball goes downward. To find the vertical distance above the court, we need to take the absolute value of the displacement:
|d_y| = 2.83 m
Therefore, the tennis ball is approximately 2.83 meters above the court when it leaves the racket.