Respuesta :
Hello,
In red : the falling volleyball h_1=14-16t^2
In Bleue : the player's jump h_2=7+24t-16t^2
If t=0 h_1=14
In red : the falling volleyball h_1=14-16t^2
In Bleue : the player's jump h_2=7+24t-16t^2
If t=0 h_1=14

Answer:
The volleyball is 14 feet above the ground at the instant the pl;ayer begins her jump.
B is correct.
Step-by-step explanation:
Here we have two situation of system of models.
[tex]\text{The height (h) of a falling volleyball as function of time (t):}h(t)=14-16t^2[/tex]
[tex]\text{The height (h) of the hands of a player as function of time (t):}h(t)=7+24t-16t^2[/tex]
We need to find the height of ball above the ground at the instant the player begins jump.
At t=0, player begins jump.
We put t=0 into [tex]h(t)=7+24t-16t^2[/tex]
Height of player hand at t=0 , h=7 feet.
Now we will set t=0 for first model.
[tex]h(0)=14-16\times 0^2\Rightarrow 14[/tex]
Thus, The volleyball is 14 feet above the ground at the instant the pl;ayer begins her jump.
B is correct.