Answer :
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.
