Answer :
Let's go through each part of the problem step-by-step.
### Part i: Write the zeroes of the given polynomial
We're given the polynomial function [tex]\( h(t) = 25t - 5t^2 \)[/tex]. The zeroes of the polynomial are the values of [tex]\( t \)[/tex] for which [tex]\( h(t) = 0 \)[/tex].
Set [tex]\( 25t - 5t^2 = 0 \)[/tex]:
[tex]\[ 25t - 5t^2 = 0 \][/tex]
[tex]\[ 5t(5 - t) = 0 \][/tex]
This equation will be satisfied when either [tex]\( 5t = 0 \)[/tex] or [tex]\( 5 - t = 0 \)[/tex].
[tex]\[ 5t = 0 \implies t = 0 \][/tex]
[tex]\[ 5 - t = 0 \implies t = 5 \][/tex]
Therefore, the zeroes of the polynomial are [tex]\( t = 0 \)[/tex] and [tex]\( t = 5 \)[/tex].
### Part ii: Find the maximum height achieved by the ball
The given function [tex]\( h(t) \)[/tex] is a quadratic equation of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -5 \)[/tex] and [tex]\( b = 25 \)[/tex]. The maximum height is achieved at the vertex of the parabola described by this quadratic equation.
The time [tex]\( t \)[/tex] at which the vertex occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{25}{2(-5)} \][/tex]
[tex]\[ t = \frac{25}{10} \][/tex]
[tex]\[ t = 2.5 \][/tex]
Now, we find the height at [tex]\( t = 2.5 \)[/tex]:
[tex]\[ h(2.5) = 25(2.5) - 5(2.5)^2 \][/tex]
[tex]\[ h(2.5) = 62.5 - 5 \cdot 6.25 \][/tex]
[tex]\[ h(2.5) = 62.5 - 31.25 \][/tex]
[tex]\[ h(2.5) = 31.25 \][/tex]
Therefore, the maximum height achieved by the ball is [tex]\( 31.25 \)[/tex] meters.
### Part iii:
iii) a) After throwing upward how much time did the ball take to reach the height of 30 m?
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) = 30 \)[/tex] meters.
Set [tex]\( 25t - 5t^2 = 30 \)[/tex]:
[tex]\[ 25t - 5t^2 = 30 \][/tex]
[tex]\[ -5t^2 + 25t - 30 = 0 \][/tex]
Divide the entire equation by -5:
[tex]\[ t^2 - 5t + 6 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( t^2 - 5t + 6 = 0 \)[/tex]. To solve it, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 6 \)[/tex].
[tex]\[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{25 - 24}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{1}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm 1}{2} \][/tex]
So, the two possible times are:
[tex]\[ t = \frac{5 + 1}{2} = 3 \][/tex]
[tex]\[ t = \frac{5 - 1}{2} = 2 \][/tex]
Therefore, the ball takes 2 seconds and 3 seconds to reach the height of 30 meters.
OR
iii) b) Find the two different values of [tex]\( t \)[/tex] when the height of the ball was 20 meters.
We need to find the times [tex]\( t \)[/tex] when the height [tex]\( h(t) = 20 \)[/tex] meters.
Set [tex]\( 25t - 5t^2 = 20 \)[/tex]:
[tex]\[ 25t - 5t^2 = 20 \][/tex]
[tex]\[ -5t^2 + 25t - 20 = 0 \][/tex]
Divide the entire equation by -5:
[tex]\[ t^2 - 5t + 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( t^2 - 5t + 4 = 0 \)[/tex]. To solve this, we can use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex].
[tex]\[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm 3}{2} \][/tex]
So, the two possible times are:
[tex]\[ t = \frac{5 + 3}{2} = 4 \][/tex]
[tex]\[ t = \frac{5 - 3}{2} = 1 \][/tex]
Therefore, the two different times when the ball reaches a height of 20 meters are 1 second and 4 seconds.
### Part i: Write the zeroes of the given polynomial
We're given the polynomial function [tex]\( h(t) = 25t - 5t^2 \)[/tex]. The zeroes of the polynomial are the values of [tex]\( t \)[/tex] for which [tex]\( h(t) = 0 \)[/tex].
Set [tex]\( 25t - 5t^2 = 0 \)[/tex]:
[tex]\[ 25t - 5t^2 = 0 \][/tex]
[tex]\[ 5t(5 - t) = 0 \][/tex]
This equation will be satisfied when either [tex]\( 5t = 0 \)[/tex] or [tex]\( 5 - t = 0 \)[/tex].
[tex]\[ 5t = 0 \implies t = 0 \][/tex]
[tex]\[ 5 - t = 0 \implies t = 5 \][/tex]
Therefore, the zeroes of the polynomial are [tex]\( t = 0 \)[/tex] and [tex]\( t = 5 \)[/tex].
### Part ii: Find the maximum height achieved by the ball
The given function [tex]\( h(t) \)[/tex] is a quadratic equation of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -5 \)[/tex] and [tex]\( b = 25 \)[/tex]. The maximum height is achieved at the vertex of the parabola described by this quadratic equation.
The time [tex]\( t \)[/tex] at which the vertex occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{25}{2(-5)} \][/tex]
[tex]\[ t = \frac{25}{10} \][/tex]
[tex]\[ t = 2.5 \][/tex]
Now, we find the height at [tex]\( t = 2.5 \)[/tex]:
[tex]\[ h(2.5) = 25(2.5) - 5(2.5)^2 \][/tex]
[tex]\[ h(2.5) = 62.5 - 5 \cdot 6.25 \][/tex]
[tex]\[ h(2.5) = 62.5 - 31.25 \][/tex]
[tex]\[ h(2.5) = 31.25 \][/tex]
Therefore, the maximum height achieved by the ball is [tex]\( 31.25 \)[/tex] meters.
### Part iii:
iii) a) After throwing upward how much time did the ball take to reach the height of 30 m?
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h(t) = 30 \)[/tex] meters.
Set [tex]\( 25t - 5t^2 = 30 \)[/tex]:
[tex]\[ 25t - 5t^2 = 30 \][/tex]
[tex]\[ -5t^2 + 25t - 30 = 0 \][/tex]
Divide the entire equation by -5:
[tex]\[ t^2 - 5t + 6 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( t^2 - 5t + 6 = 0 \)[/tex]. To solve it, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 6 \)[/tex].
[tex]\[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{25 - 24}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{1}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm 1}{2} \][/tex]
So, the two possible times are:
[tex]\[ t = \frac{5 + 1}{2} = 3 \][/tex]
[tex]\[ t = \frac{5 - 1}{2} = 2 \][/tex]
Therefore, the ball takes 2 seconds and 3 seconds to reach the height of 30 meters.
OR
iii) b) Find the two different values of [tex]\( t \)[/tex] when the height of the ball was 20 meters.
We need to find the times [tex]\( t \)[/tex] when the height [tex]\( h(t) = 20 \)[/tex] meters.
Set [tex]\( 25t - 5t^2 = 20 \)[/tex]:
[tex]\[ 25t - 5t^2 = 20 \][/tex]
[tex]\[ -5t^2 + 25t - 20 = 0 \][/tex]
Divide the entire equation by -5:
[tex]\[ t^2 - 5t + 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( t^2 - 5t + 4 = 0 \)[/tex]. To solve this, we can use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 4 \)[/tex].
[tex]\[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{25 - 16}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm \sqrt{9}}{2} \][/tex]
[tex]\[ t = \frac{5 \pm 3}{2} \][/tex]
So, the two possible times are:
[tex]\[ t = \frac{5 + 3}{2} = 4 \][/tex]
[tex]\[ t = \frac{5 - 3}{2} = 1 \][/tex]
Therefore, the two different times when the ball reaches a height of 20 meters are 1 second and 4 seconds.
