A volleyball court measures 30-feet wide by 60-feet long. The net is located 30 feet from the serve line. Assume the server stands 31 feet from the net when she serves the ball. The net is approximately 7.5 feet from the ground. Write an equation that would show the path of a served ball that would clear the net and explain how you found your equation. Then, document your equation.
Help I will give 20 pts

The ball follows a parabolic path, where the equation that models the path of the served ball is:

The horizontal displacement is represented as:

Where:

Make t the subject

Substitute 31 for x

The vertical displacement is represented as:

Where y = 7.5.

So, we have:

Make h the subject

Substitute

Cancel out v

Rewrite as:

Hence, the equation that models the path of the served ball is:

Read more about displacement at:

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This is a tough question to answer, not because it's hard, but because it's hard to know what's already been taught.  For example we can express the components of the velocity off the racket in polar or rectangular form.  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 v₀.  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.  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.  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.

step-by-step explanation:

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9