This is a tough question to answer, not because it's hard, but because it's hard to know what's already been taught. Â For example we can express the components of the velocity off the racket in polar or rectangular form. Â We could write the equation to include air resistance, definitely a concern for a real tennis serve, but not usually for high school math.
It's a projectile; there are really two variables, the initial velocity off the racket in the horizontal direction, and the initial velocity of the racket in the vertical direction.
Let's say the server hits the ball at her usual service velocity vâ‚€. Â Let's say the ball is at height h when it comes in contact with the racket. We'll further assume she gets to control the angle of elevation A she hits the ball.
In the horizontal direction we have constant velocity.
x = ( v₀ cos θ ) t
For us, we're interested in the time the ball takes to travel 31 horizontal feet. Â We set x=31 feet.
t = x /  ( v₀ cos θ )  = 31 /  ( v₀ cos θ )
In the y direction we have motion in a constant gravitational field:
y = -16t² + ( v₀ sin θ ) t + h
-16 (feet per second per second) is the acceleration of gravity.  v₀ sin θ  is the vertical component of velocity.  h is the initial height of the ball when it hit the racket.
We need y > 7.5 to clear the net. Â Let's set it equal to 7.5 to calculate the 'just barely' numbers.
7.5 = -16t² + ( v₀ sin θ ) t + h
We substitute  t = 31 /  ( v₀ cos θ )
7.5 - h = -16(31 /  ( v₀ cos θ ))² + ( v₀ sin θ ) ( 31 /  ( v₀ cos θ ) )
7.5 - h = -16(31²) /( v₀² cos²θ ) + 31 v₀ tan θ
Let's stop here; that's our equation, we don't need to solve it.  If we fixed h and v₀ this would be a trig problem to solve for θ, the angle she needs to hit the ball given her height and strength.
I've been documenting along the way; to recap, 7.5 is the height in feet of the net, h is the height of the ball when hit, -16 is the acceleration of gravity, 31 is the horizontal distance to the net, v₀ is the speed of the ball off the racket, θ is the angle of elevation of the ball off the racket.
The ball follows a parabolic path, where the equation that models the path of the served ball is: [tex]\mathbf{h = 7.5 + 16(\frac{31}{vcos\theta})^2 - 31 tan\theta }[/tex]
The horizontal displacement is represented as:
[tex]\mathbf{x = (v_0 cos \theta ) t}[/tex]
Where:
[tex]\mathbf{v \to velocity}[/tex]
[tex]\mathbf{x = 31 \to horizontal\ displacement}[/tex]
[tex]\mathbf{t \to time}[/tex]
Make t the subject
[tex]\mathbf{t = \frac{x}{v_0 cos\theta}}[/tex]
Substitute 31 for x
[tex]\mathbf{t = \frac{31}{v cos\theta}}[/tex]
The vertical displacement is represented as:
[tex]\mathbf{y = -16t^2 + ( v sin \theta ) t + h}[/tex]
Where y = 7.5.
So, we have:
[tex]\mathbf{7.5 = -16t^2 + ( v sin \theta ) t + h}[/tex]
Make h the subject
[tex]\mathbf{h = 7.5 + 16t^2 - ( v sin \theta ) t }[/tex]
Substitute [tex]\mathbf{t = \frac{31}{v cos\theta}}[/tex]
[tex]\mathbf{h = 7.5 + 16 \times (\frac{31}{vcos\theta})^2 - ( v sin \theta ) \times (\frac{31}{vcos\theta}) }[/tex]
[tex]\mathbf{h = 7.5 + 16 \times (\frac{31}{vcos\theta})^2 - \frac{31v sin\theta}{vcos\theta} }[/tex]
Cancel out v
[tex]\mathbf{h = 7.5 + 16 \times (\frac{31}{vcos\theta})^2 - \frac{31 sin\theta}{cos\theta} }[/tex]
Rewrite as:
[tex]\mathbf{h = 7.5 + 16 \times (\frac{31}{vcos\theta})^2 - 31 tan\theta }[/tex]
[tex]\mathbf{h = 7.5 + 16(\frac{31}{vcos\theta})^2 - 31 tan\theta }[/tex]
Hence, the equation that models the path of the served ball is:
[tex]\mathbf{h = 7.5 + 16(\frac{31}{vcos\theta})^2 - 31 tan\theta }[/tex]
Read more about displacement at:
https://brainly.com/question/24654533
