A tennis ball is struck and departs from the racket horizontally with a speed of 27.4 m/s. The ball hits the court at a horizontal distance of 20.0 from the racket. How far above the court is the tennis ball when it leaves the racket?
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To find the vertical distance above the court at which the tennis ball leaves the racket, we first need to determine the time it takes for the ball to hit the ground.
We can start by using the horizontal distance traveled by the ball and its horizontal velocity to calculate the time of flight.
The formula for horizontal distance is given by:
horizontal distance = horizontal velocity * time
In this case, the horizontal distance is 20.0 m and the horizontal velocity is 27.4 m/s. So, we have:
20.0 m = 27.4 m/s * time
Now, we can solve for time:
time = horizontal distance / horizontal velocity
= 20.0 m / 27.4 m/s
≈ 0.73 s
So, it takes approximately 0.73 seconds for the tennis ball to hit the ground.
Now, we can use this time and the vertical motion equations to find the vertical distance the ball traveled.
The vertical motion equation is given by:
vertical distance = initial vertical velocity * time + (1/2) * acceleration * time^2
Since the ball is leaving the racket horizontally, the initial vertical velocity is 0 m/s. The acceleration due to gravity is approximately 9.8 m/s^2.
Substituting these values into the equation, we get:
vertical distance = 0 m/s * 0.73 s + (1/2) * 9.8 m/s^2 * (0.73 s)^2
= 0 + (1/2) * 9.8 m/s^2 * 0.5329 s^2
≈ 2.61 m
Therefore, the tennis ball is approximately 2.61 meters above the court when it leaves the racket.