Question

The height y (in feet) of a ball thrown by a child is

Answer 1

Answer:

1. 3 ft
2. 129 ft
3. 84.5 ft

Step-by-step explanation:

The given equation,

$y=-\dfrac{1}{14}x^2+6x+3$

where,

x is the horizontal distance in feet and

y is the height of the ball in feet.

How high is the ball when it leaves the child's hand.

At the beginning when the ball leaves the hand of the child, the horizontal distance was 0, so putting x=0 we can get the height of the ball.

$y_0=-\dfrac{1}{14}(0)^2+6(0)+3=3\ ft$

What is the maximum height of the ball.

As the quadratic function has negative leading coefficient, so it will open downwards and at the vertex we will get the maximum value of the function.

The coordinates of vertex are,

$\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)$

Putting the values,

$=\left(-\dfrac{6}{2\times -\frac{1}{14}},f\left(-\dfrac{6}{2\times -\frac{1}{14}}\right)\right)$

$=\left(42,f\left(42\right)\right)$

Now,

$y_{max}=f(42)=-\dfrac{1}{14}(42)^2+6(42)+3= 129\ ft$

How far from the child does the ball strike the ground.

At the end i.e when the ball touches the ground, the height of the ball is 0. So putting the value of y=0, in the equation,

$\Rightarrow 0=-\dfrac{1}{14}x^2+6x+3$

$\Rightarrow -\dfrac{1}{14}x^2+6x+3=0$

$\Rightarrow -x^2+6x(14)+3(14)=0$

$\Rightarrow -x^2+84x+42=0$

Solving the quadratic equation, we get

$\Rightarrow x=-0.497,84.5$

As distance can not be negative, so only considering positive value,

$x=84.5\ ft$