Answer :
Sure, let's solve this in a step-by-step manner.
Given:
The height of the rocket after [tex]\( t \)[/tex] seconds is given by the quadratic equation:
[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 0. Therefore, we need to find the time [tex]\( t \)[/tex] when [tex]\( y = 0 \)[/tex].
1. Set the height equation to zero:
[tex]\[ -16t^2 + 76t + 8 = 0 \][/tex]
2. Solve this quadratic equation for [tex]\( t \)[/tex]. The solutions to the quadratic equation (using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]) can be given in the form:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
3. In our equation, [tex]\( a = -16 \)[/tex], [tex]\( b = 76 \)[/tex], and [tex]\( c = 8 \)[/tex]. So, we solve for [tex]\( t \)[/tex]:
After solving, we get the two solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{19}{8} - \frac{\sqrt{393}}{8} \][/tex]
[tex]\[ t_2 = \frac{19}{8} + \frac{\sqrt{393}}{8} \][/tex]
These are the two times at which the rocket's height is zero (i.e., it is at ground level). For negative values, we discard the solution as the time can't be negative.
### Part (b): How high is the maximum height of the rocket?
The maximum height of a projectile represented by a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the vertex of the parabola. The time at which the maximum height occurs (vertex) is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
1. In our equation, [tex]\( a = -16 \)[/tex] and [tex]\( b = 76 \)[/tex].
2. So the time [tex]\( t \)[/tex] at which maximum height is achieved:
[tex]\[ t = -\frac{76}{2 \cdot (-16)} = \frac{76}{32} = \frac{19}{8} \][/tex]
3. Substitute [tex]\( t = \frac{19}{8} \)[/tex] back into the height equation to find the maximum height:
[tex]\[ y = -16 \left( \frac{19}{8} \right)^2 + 76 \left( \frac{19}{8} \right) + 8 \][/tex]
After evaluating, we get the maximum height of the rocket as:
[tex]\[ y = 98.25 \][/tex]
### Summary:
a) The rocket hits the ground at times:
[tex]\[ t_1 = \frac{19}{8} - \frac{\sqrt{393}}{8} \quad \text{and} \quad t_2 = \frac{19}{8} + \frac{\sqrt{393}}{8} \][/tex]
b) The maximum height of the rocket is:
[tex]\[ y = 98.25 \][/tex]
So, the time at which the rocket hits the ground and the maximum height has been accurately determined.
Given:
The height of the rocket after [tex]\( t \)[/tex] seconds is given by the quadratic equation:
[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 0. Therefore, we need to find the time [tex]\( t \)[/tex] when [tex]\( y = 0 \)[/tex].
1. Set the height equation to zero:
[tex]\[ -16t^2 + 76t + 8 = 0 \][/tex]
2. Solve this quadratic equation for [tex]\( t \)[/tex]. The solutions to the quadratic equation (using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]) can be given in the form:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
3. In our equation, [tex]\( a = -16 \)[/tex], [tex]\( b = 76 \)[/tex], and [tex]\( c = 8 \)[/tex]. So, we solve for [tex]\( t \)[/tex]:
After solving, we get the two solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{19}{8} - \frac{\sqrt{393}}{8} \][/tex]
[tex]\[ t_2 = \frac{19}{8} + \frac{\sqrt{393}}{8} \][/tex]
These are the two times at which the rocket's height is zero (i.e., it is at ground level). For negative values, we discard the solution as the time can't be negative.
### Part (b): How high is the maximum height of the rocket?
The maximum height of a projectile represented by a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] occurs at the vertex of the parabola. The time at which the maximum height occurs (vertex) is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
1. In our equation, [tex]\( a = -16 \)[/tex] and [tex]\( b = 76 \)[/tex].
2. So the time [tex]\( t \)[/tex] at which maximum height is achieved:
[tex]\[ t = -\frac{76}{2 \cdot (-16)} = \frac{76}{32} = \frac{19}{8} \][/tex]
3. Substitute [tex]\( t = \frac{19}{8} \)[/tex] back into the height equation to find the maximum height:
[tex]\[ y = -16 \left( \frac{19}{8} \right)^2 + 76 \left( \frac{19}{8} \right) + 8 \][/tex]
After evaluating, we get the maximum height of the rocket as:
[tex]\[ y = 98.25 \][/tex]
### Summary:
a) The rocket hits the ground at times:
[tex]\[ t_1 = \frac{19}{8} - \frac{\sqrt{393}}{8} \quad \text{and} \quad t_2 = \frac{19}{8} + \frac{\sqrt{393}}{8} \][/tex]
b) The maximum height of the rocket is:
[tex]\[ y = 98.25 \][/tex]
So, the time at which the rocket hits the ground and the maximum height has been accurately determined.
