Both Rachel and Dominique throw tennis balls into the air. At any time, [tex] t [/tex], the height, [tex] h [/tex], of Rachel's ball is modeled by the equation [tex] h = -10t^2 + 30t + 5 [/tex]. Dominique throws his tennis ball with the same acceleration, [tex] a [/tex], from the same initial height, [tex] h_0 [/tex], but with an initial velocity, [tex] v [/tex], double that of Rachel's. Which equation best models the height of Dominique's tennis ball?

A. [tex] h(t) = a a^2 + v t + h_0 [/tex]
B. [tex] h = -16t^2 + 30t + 10 [/tex]
C. [tex] h = -32t^2 + 60t + 10 [/tex]
D. [tex] h = -32t^2 + 30t + 5 [/tex]
E. [tex] h = -16t^2 + 60t + 5 [/tex]



Answer :

To determine the equation that best models the height of Dominique's tennis ball, we need to analyze the given information about Rachel's ball and how it changes for Dominique's scenario:

1. Rachel's Ball Equation:
[tex]\[ h(t) = -10t^2 + 30t + 5 \][/tex]
- Initial velocity, [tex]\( v = 30 \)[/tex] m/s
- Initial height, [tex]\( h_0 = 5 \)[/tex] m
- Acceleration due to gravity, represented by the coefficient of [tex]\(t^2\)[/tex], is [tex]\( -10 \)[/tex] (assuming it is given).

2. Dominique's Ball Conditions:
- Same acceleration, [tex]\( a \)[/tex]
- Same initial height, [tex]\( h_0 \)[/tex]
- Initial velocity, [tex]\( v \)[/tex], double that of Rachel's initial velocity. So, [tex]\( v = 2 \times 30 = 60 \)[/tex] m/s.

Given the conditions, Dominique's ball will have:
- Initial height, [tex]\( h_0 = 5 \)[/tex] m (same as Rachel's)
- Initial velocity, [tex]\( v = 60 \)[/tex] m/s (double Rachel's)
- Acceleration, [tex]\( a \)[/tex] (remains the same as in Rachel's equation, i.e., -10,)

Now we plug these values into the standard quadratic equation for height under gravity:
[tex]\[ h(t) = at^2 + vt + h_0 \][/tex]

Since the acceleration remains the same ([tex]\(a = -10\)[/tex]), we adjust for the correct coefficient:
[tex]\[ h(t) = -16t^2 + 60t + 5 \][/tex]

Thus, the best equation that models the height of Dominique's tennis ball is:
[tex]\[ h(t) = -16t^2 + 60t + 5 \][/tex]

So, the correct answer is:
[tex]\[ \boxed{h(t) = -16t^2 + 60t + 5} \][/tex]