Tazmin Horton

2021-08-12

The height of the ball increases until it reaches a maximum height of 8 yards, 20 yards away from the player.
To find: Which kick travels further before hitting the ground.

wheezym

Step 1
Which kick travels higher.
Here the maximum height of the first ball is 8 yards.
When the ball reached its maximum height it was 20 yards away
from player, so the first ball travels $20\left(2\right)=40$ yards
To find the maximum height find the value of vertex.
If the quadratic equation is of the form $a{x}^{2}+bx+c=0$ then the vertex for the equation
$a{x}^{2}+bx+c=0$ is $x=\frac{-b}{2a}$
Step 2
Given that a second kick is modelled by,
$y=x\left(0.4-0.008x\right)$
After solving,
$y=0.4x-0.008{x}^{2}$
Compare $y=0.4x-0.008{x}^{2}$ with $a{x}^{2}+bx+c$ get,

So the vertex is,
$x=\frac{-b}{a}$
$=\frac{-0.4}{-0.008}$
$=\frac{0.4}{0.008}$
$=25$
$⇒x=25$
Substitute $x=25$ in $y=x\left(0.4-0.008x\right)$,
$y=25\left(0.4-0.008\left(25\right)\right)$
$y=25\left(0.4-0.2\right)$
$y=5$
So the maximum height of the second ball is 5 yards and the second ball travels
$25\left(2\right)=50$ yards.
Step 3
Therefore,
The maximum height of the second ball is 5 yards and the second ball travels $25\left(2\right)=50$ yards.

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