Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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]
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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.