A tennis ball is struck and departs from the racket horizontally with a speed of 29.1 m/s. The ball hits the court at a horizontal distance of 19.0 from the racket. How far above the court is the tennis ball when it leaves the racket?

2.089m

To determine the height above the court at which the tennis ball leaves the racket, we can use the equation for projectile motion.

The horizontal distance covered by the ball, x, can be calculated using the equation:

x = v₀x * t

where:
- x = horizontal distance (19.0 m)
- v₀x = initial horizontal velocity (29.1 m/s)
- t = time of flight

Since the initial velocity in the horizontal direction is constant and there is no acceleration in that direction, the time of flight is given by:

t = x / v₀x

Substituting the given values, we have:

t = 19.0 m / 29.1 m/s

Now, we need to find the vertical distance or height above the court. The equation for vertical displacement in projectile motion is:

y = v₀y * t + (1/2) * a * t²

where:
- y = vertical displacement or height above the court
- v₀y = initial vertical velocity (unknown)
- a = acceleration due to gravity (-9.8 m/s²)

The initial vertical velocity, v₀y, is zero since the ball starts at its highest point and falls under the influence of gravity. Thus, the equation becomes:

y = (1/2) * a * t²

Substituting the values for acceleration due to gravity, and the previously calculated value for time of flight, we have:

y = (1/2) * (-9.8 m/s²) * (19.0 m / 29.1 m/s)²

Evaluating this expression will give us the height above the court that the ball leaves the racket.

To find the height above the court at which the tennis ball leaves the racket, we need to analyze the projectile motion of the ball.

We can split the ball's motion into two components: horizontal and vertical.

1. Horizontal motion:
The ball is struck horizontally, so there is no acceleration in the horizontal direction. The horizontal distance traveled by the ball is given as 19.0 m.

2. Vertical motion:
The ball is subject to the acceleration due to gravity, which causes it to move vertically. We need to find the vertical distance or height.

To solve for the height above the court, we can use the kinematic equation:

h = u * t + (1/2) * g * t^2

where:
h = height (vertical distance)
u = initial vertical velocity (0 m/s, as there is no vertical velocity initially)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of flight

To find the time of flight (t), we can use the equation:

d = u * t + (1/2) * a * t^2

where:
d = horizontal distance traveled
u = initial horizontal velocity (29.1 m/s)
a = acceleration in the horizontal direction (0 m/s^2, as there is no horizontal acceleration)

In this case, we want to find the time of flight, so we rearrange the equation:

t = (d - u * t) / (1/2 * a * t)

Plugging in the values we know:
d = 19.0 m
u = 29.1 m/s
a = 0 m/s^2

t = (19.0 m - 29.1 m/s * t) / (1/2 * 0 m/s^2 * t)

Simplifying the equation:
t = (19.0 m) / (29.1 m/s)

Solving for t:
t ≈ 0.653 seconds

Now that we have the time of flight, we can substitute it back into the vertical motion equation:

h = 0 m/s * 0.653 s + (1/2) * 9.8 m/s^2 * (0.653 s)^2

Simplifying the equation:
h = 0 m + 3.188 m

Therefore, the tennis ball is approximately 3.188 meters above the court when it leaves the racket.

time = distance/speed

t = 19m / 29.1m/s = .653s

So, how far does it fall in .653 seconds?