A tennis ball is struck and departs from the racket horizontally with a speed of 27.7 m/s. The ball hits the court at a horizontal distance of 20.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?
Dx = Vx * Tf = 20.6 m.
27.7 * Tf = 20.6
Tf = 0.744 = Fall time.
h = 0.5g*Tf^2
g = +9.8 m/s^2
Solve for h.
To find the height above the court at which the tennis ball leaves the racket, we will use the principles of projectile motion.
We know that the horizontal distance traveled by the ball is 20.6 m. This is the same as the horizontal velocity (since there is no acceleration in the horizontal direction) multiplied by the time of flight.
So, let's first find the time of flight. We can use the equation:
distance = velocity * time.
In this case, the distance is 20.6 m and the velocity is 27.7 m/s. Solving for time:
time = distance / velocity = 20.6 m / 27.7 m/s.
Calculating this, we find the time of flight to be approximately 0.744 seconds.
Now, since the ball is launched horizontally, the vertical motion is influenced only by the acceleration due to gravity (9.8 m/s²) acting downward. We can use the equation:
height = initial vertical velocity * time + (1/2) * acceleration * time².
In this case, the initial vertical velocity is 0 because the ball is not launched vertically. Therefore, the equation simplifies to:
height = (1/2) * acceleration * time².
Substituting the values:
height = (1/2) * (9.8 m/s²) * (0.744 seconds)².
Evaluating this expression, we find that the height above the court at which the tennis ball leaves the racket is approximately 2.74 meters.