As a tennis ball is struck, it departs from the racket horizontally with a speed of 27.1 m/s. The ball hits the court at a horizontal distance of 20.0 m from the racket. How far above the court is the tennis ball when it leaves the racket
Dx = Vx * Tf = 20 m.
27.1 * Tf = 20
Tf = 0.738 s. = Fall time.
h = Yo*Tf + 0.5g*Tf^2
h = 0 + 4.9*0.738^2 = 2.67 m.
To find the height at which the tennis ball leaves the racket, we can use the horizontal distance and initial horizontal velocity of the ball.
We can use the formula:
distance = velocity × time
In this case, the distance is given as 20.0 m and the horizontal velocity as 27.1 m/s.
Since there is no horizontal acceleration, the time taken by the ball to travel 20.0 m horizontally is the same as the time taken for it to fall vertically from its initial height to the ground.
The time taken for the ball to fall can be calculated using the vertical motion equation:
distance = (1/2) × acceleration × time^2
In this case, the distance fallen is the same as the initial height above the court (which we want to find). The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s^2.
We can rearrange the equation to solve for time:
time = sqrt((2 × distance) / acceleration)
Using the given values, we can substitute them into the equation to find the time taken for the ball to fall.
time = sqrt((2 × height) / 9.8)
Now, we can equate the time taken to the time calculated from the horizontal motion:
time = distance / velocity
Substituting the given values and solving for height:
sqrt((2 × height) / 9.8) = 20.0 / 27.1
Simplifying the equation:
2 × height = (20.0 / 27.1)^2 × 9.8
Now we can solve for the height:
height = [(20.0 / 27.1)^2 × 9.8] / 2
Calculating this expression:
height ≈ 2.59 meters
Therefore, the tennis ball is approximately 2.59 meters above the court when it leaves the racket.