A tennis ball is struck and departs from the racket horizontally with a speed of 26.7 m/s. The ball hits the court at a horizontal distance of 18.7 from the racket. How far above the court is the tennis ball when it leaves the racket?
To calculate how far above the court the tennis ball is when it leaves the racket, we can use the equations of motion.
First, we need to find the time it takes for the ball to travel a horizontal distance of 18.7 m. We can use the equation:
distance = speed * time
Rearranging the equation to solve for time:
time = distance / speed
Substituting the values given, we have:
time = 18.7 m / 26.7 m/s
time = 0.699 sec (approximately)
Now, since the ball is projected horizontally, there is no initial vertical velocity. Therefore, we can use the equation of motion for vertical motion:
vertical distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the ball is hitting the court, the final vertical distance is 0 (the height of the court). Thus, the equation becomes:
0 = (initial velocity * time) + (0.5 * acceleration * time^2)
The acceleration in this case is due to gravity and is equal to -9.8 m/s^2 (taking the downward direction as positive).
Substituting the values, the equation becomes:
0 = (0 * 0.699 sec) + (0.5 * (-9.8 m/s^2) * (0.699 sec)^2)
By simplifying the equation, we find that:
0.5 * (-9.8 m/s^2) * (0.699 sec)^2 = (0.5 * (-9.8 m/s^2) * (0.488601 sec^2))
= -2.38348 m
Therefore, the tennis ball is approximately 2.38 meters above the court when it leaves the racket.