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?
54m
To find out how far above the court the tennis ball is when it leaves the racket, we can use the equation for horizontal motion.
First, let's assume that the time taken by the tennis ball to hit the court is t seconds.
The horizontal distance traveled by the ball (d) can be calculated using the equation:
d = v * t
where:
d = horizontal distance traveled by the ball (18.4 m)
v = horizontal velocity of the ball (26.4 m/s)
t = time taken by the ball to hit the court (unknown)
Rearranging the equation, we get:
t = d / v
Substituting the given values, we have:
t = 18.4 m / 26.4 m/s
t ≈ 0.697 seconds
Now, let's consider the vertical motion of the ball. We can use the equation for vertical motion to determine the height above the court.
The vertical distance or the height above the court (h) can be calculated using the equation:
h = (1/2) * g * t^2
where:
h = vertical distance or height above the court (unknown)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken by the ball to hit the court (0.697 seconds)
Substituting the values, we have:
h = (1/2) * 9.8 m/s^2 * (0.697 s)^2
h ≈ 2.869 meters
Therefore, the tennis ball is approximately 2.869 meters above the court when it leaves the racket.