Question

select the correct option from each drop - down menu. anna throws a ball in the air from her balcony. the following equation models the height of the ball, in meters, after t seconds: $h(t)=-5t^{2}+9t + 80$. anna wants to rewrite the equation to determine after how many seconds the ball hits the ground. use this situation to complete the following statements. the function drop - down menu reveals that the balls drop - down menu above the ground will be drop - down menu meters after drop - down menu seconds.

Explanation:

To determine when the ball hits the ground, we set \( h(t) = 0 \) (since height is 0 when it hits the ground) and solve the quadratic equation \( -5t^2 + 9t + 80 = 0 \).

Step 1: Set up the equation for ground - hitting

When the ball hits the ground, its height \( h(t)=0 \). So we have the equation:
\[ - 5t^{2}+9t + 80=0\]
We can multiply both sides by - 1 to make the coefficient of \( t^{2}\) positive:
\[5t^{2}-9t - 80 = 0\]

Step 2: Solve the quadratic equation \( ax^{2}+bx + c = 0 \) (here \( x = t\), \( a = 5\), \( b=-9\), \( c = - 80\))

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)
First, calculate the discriminant \( D=b^{2}-4ac\)
\[D=(-9)^{2}-4\times5\times(-80)=81 + 1600=1681\]
Then,
\[t=\frac{9\pm\sqrt{1681}}{2\times5}=\frac{9\pm41}{10}\]
We have two solutions:
\(t_1=\frac{9 + 41}{10}=\frac{50}{10} = 5\)
\(t_2=\frac{9-41}{10}=\frac{-32}{10}=-3.2\)
Since time cannot be negative in this context, we take \( t = 5\) seconds.

When \( t = 5\), we can check the height function \( h(5)=-5\times(5)^{2}+9\times5 + 80=-125 + 45+80 = 0\)

So the function \( h(t)=-5t^{2}+9t + 80\) reveals that the ball's height above the ground will be \( 0\) meters after \( 5\) seconds.

If we assume the first drop - down is for the function (which is \( h(t)=-5t^{2}+9t + 80\)), the second is for "height", the third is for "0", and the fourth is for "5".

Answer:

The function \(\boldsymbol{h(t)=-5t^{2}+9t + 80}\) reveals that the ball's \(\boldsymbol{\text{height}}\) above the ground will be \(\boldsymbol{0}\) meters after \(\boldsymbol{5}\) seconds.