As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.2 m/s. The ball hits the court at a horizontal distance of 19.5 m from the racket. How far above the court is the tennis ball when it leaves the racket?
v=28.2 m/s,
L=19.5 m.
L=v t =>
t=L/v=...
h= gt²/2= ...
To find the height at which the tennis ball leaves the racket, we can use the kinematic equation for horizontal motion.
The horizontal distance traveled by the ball can be given by:
d = v * t
where d is the horizontal distance, v is the horizontal velocity, and t is the time of flight.
In this case, the horizontal distance traveled is 19.5 m and the horizontal velocity is 28.2 m/s.
19.5 m = 28.2 m/s * t
Now we need to find the time of flight. Since we know the horizontal velocity, we can use it to find the time it takes for the ball to travel 19.5 m horizontally.
t = d / v
t = 19.5 m / 28.2 m/s
Now that we have the time of flight, we can find the height of the ball using the vertical motion equation:
h = v0 * t + (1/2) * g * t^2
where h is the height, v0 is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the initial vertical velocity is 0 m/s since the ball is struck horizontally. Therefore, the equation simplifies to:
h = (1/2) * g * t^2
Now we can substitute the values we have and calculate the height:
h = (1/2) * 9.8 m/s^2 * (19.5 m / 28.2 m/s)^2
h ≈ 3.24 m
Therefore, the tennis ball is approximately 3.24 meters above the court when it leaves the racket.