Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Sure, let's tackle the problem step-by-step.
#### Given:
The equation for the height [tex]\( y \)[/tex] of the rocket after [tex]\( t \)[/tex] seconds is given by:
[tex]\[ y = -16t^2 + 76t + 8 \][/tex]
### Part a) When would the rocket hit the ground?
The rocket hits the ground when its height [tex]\( y \)[/tex] is zero. Therefore, we need to solve the equation:
[tex]\[ 0 = -16t^2 + 76t + 8 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
[tex]\[ a = -16, \quad b = 76, \quad c = 8 \][/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values, we get:
[tex]\[ t = \frac{-76 \pm \sqrt{76^2 - 4(-16)(8)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-76 \pm \sqrt{5776 + 512}}{-32} \][/tex]
[tex]\[ t = \frac{-76 \pm \sqrt{6288}}{-32} \][/tex]
[tex]\[ t = \frac{-76 \pm 79.32}{-32} \][/tex]
Thus, we get two solutions:
[tex]\[ t_1 = \frac{-76 + 79.32}{-32} \approx \frac{3.32}{-32} \approx -0.104 \quad \text{(Not a valid solution as time can't be negative)} \][/tex]
[tex]\[ t_2 = \frac{-76 - 79.32}{-32} \approx \frac{-155.32}{-32} \approx 4.85 \][/tex]
Therefore, the rocket hits the ground after approximately [tex]\( t = 4.85 \)[/tex] seconds.
### Part b) How high is the maximum height of the rocket?
The maximum height of a parabolic trajectory in the form [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the vertex, given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here:
[tex]\[ a = -16 \quad \text{and} \quad b = 76 \][/tex]
Therefore:
[tex]\[ t = -\frac{76}{2(-16)} = \frac{76}{32} = 2.375 \text{ seconds} \][/tex]
To find the maximum height, we substitute [tex]\( t = 2.375 \)[/tex] back into the original equation:
[tex]\[ y = -16(2.375)^2 + 76(2.375) + 8 \][/tex]
[tex]\[ y = -16(5.640625) + 180.5 + 8 \][/tex]
[tex]\[ y = -90.25 + 180.5 + 8 \][/tex]
[tex]\[ y = 98.25 \][/tex]
Therefore, the maximum height of the rocket is [tex]\( 98.25 \)[/tex] feet.
### Summary:
a) The rocket hits the ground after approximately [tex]\( 4.85 \)[/tex] seconds.
b) The maximum height of the rocket is [tex]\( 98.25 \)[/tex] feet.
#### Given:
The equation for the height [tex]\( y \)[/tex] of the rocket after [tex]\( t \)[/tex] seconds is given by:
[tex]\[ y = -16t^2 + 76t + 8 \][/tex]
### Part a) When would the rocket hit the ground?
The rocket hits the ground when its height [tex]\( y \)[/tex] is zero. Therefore, we need to solve the equation:
[tex]\[ 0 = -16t^2 + 76t + 8 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
[tex]\[ a = -16, \quad b = 76, \quad c = 8 \][/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values, we get:
[tex]\[ t = \frac{-76 \pm \sqrt{76^2 - 4(-16)(8)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-76 \pm \sqrt{5776 + 512}}{-32} \][/tex]
[tex]\[ t = \frac{-76 \pm \sqrt{6288}}{-32} \][/tex]
[tex]\[ t = \frac{-76 \pm 79.32}{-32} \][/tex]
Thus, we get two solutions:
[tex]\[ t_1 = \frac{-76 + 79.32}{-32} \approx \frac{3.32}{-32} \approx -0.104 \quad \text{(Not a valid solution as time can't be negative)} \][/tex]
[tex]\[ t_2 = \frac{-76 - 79.32}{-32} \approx \frac{-155.32}{-32} \approx 4.85 \][/tex]
Therefore, the rocket hits the ground after approximately [tex]\( t = 4.85 \)[/tex] seconds.
### Part b) How high is the maximum height of the rocket?
The maximum height of a parabolic trajectory in the form [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the vertex, given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here:
[tex]\[ a = -16 \quad \text{and} \quad b = 76 \][/tex]
Therefore:
[tex]\[ t = -\frac{76}{2(-16)} = \frac{76}{32} = 2.375 \text{ seconds} \][/tex]
To find the maximum height, we substitute [tex]\( t = 2.375 \)[/tex] back into the original equation:
[tex]\[ y = -16(2.375)^2 + 76(2.375) + 8 \][/tex]
[tex]\[ y = -16(5.640625) + 180.5 + 8 \][/tex]
[tex]\[ y = -90.25 + 180.5 + 8 \][/tex]
[tex]\[ y = 98.25 \][/tex]
Therefore, the maximum height of the rocket is [tex]\( 98.25 \)[/tex] feet.
### Summary:
a) The rocket hits the ground after approximately [tex]\( 4.85 \)[/tex] seconds.
b) The maximum height of the rocket is [tex]\( 98.25 \)[/tex] feet.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
