A tennis ball is struck and departs from the racket horizontally with a speed of 26.4 m/s. The ball hits the court at a horizontal distance of 18.4 from the racket. How far above the court is the tennis ball when it leaves the racket?
Xo = 26.4 m/s.
Dx = 18.4 m.
Tf = Fall time.
Dx = Xo*Tf = 26.4 * Tf = 18.4 m.
Tf = 0.697 s.
h = 0.5g*Tf^2 = 4.9*0.697^2 = 2.38 m.
To find the height at which the tennis ball leaves the racket, we can use the equations of motion.
We know the initial velocity of the ball in the horizontal direction is 26.4 m/s, and we can assume there is no initial vertical velocity. We also know the horizontal distance traveled by the ball before hitting the ground is 18.4 m.
Let's assume that the height of the ball when it leaves the racket is "h" meters.
In the horizontal direction:
distance = speed × time
Since there is no horizontal acceleration, the time taken by the ball to travel 18.4 m horizontally is the same as the time taken to leave the racket.
So, 18.4 = 26.4 × time
Simplifying the equation, we have:
time = 18.4 / 26.4
Now, we can use the equations of motion in the vertical direction to find the height "h."
The equation relating distance, initial velocity, time, and acceleration in the vertical direction is:
distance = initial velocity × time + (1/2) × acceleration × time²
In this case, the initial vertical velocity is 0 (assuming no initial vertical velocity), and the acceleration is due to gravity, which is approximately -9.8 m/s² (negative because it acts downwards).
So, the equation becomes:
h = 0 × time + (1/2) × (-9.8) × time²
Simplifying the equation, we have:
h = -4.9 × time²
Now, we substitute the value of "time" we found earlier:
h = -4.9 × (18.4 / 26.4)²
Evaluating the expression, we find:
h ≈ -4.9 × (0.697)²
h ≈ -4.9 × 0.485
h ≈ -2.3725
Since height cannot be negative, we take the absolute value of the answer:
h ≈ 2.3725
Therefore, the tennis ball is approximately 2.37 meters above the court when it leaves the racket.