Question

A person standing close to the edge on top of a 96 foot building throws a ball vertically upward. The quadratic function h= - 16t^2 + 80t + 96 models the balls height above the ground, h, in feet, t seconds after it was thrown. A. What is the maximum height of the ball ___ ft B. How many seconds does it take until the ball hits the ground? ___ seconds

Answer 1

Explanation

The height of the ball as a function of the time is given by the following formula:

$h(t)=-16t^2+80t+96.$

A. Maximum height of the ball

To find the value of the maximum height, we maximum the function h(t) computing its first derivative and equalling the result to zero:

$h^{\prime}(t)=-2*16t+80=0\Rightarrow-32t+80=0$

Solving the last equation for t, we get time t when the height is maximum:

$\begin{gathered} 32t=80, \\ t=\frac{80}{32}=2.5. \end{gathered}$

So the maximum height is given by the value of h(t) when t = 2.5:

$h(2.5)=-16*2.5^2+80*2.5+96=196.$

The maximum height is 196ft.

B. Time until the ball hits the ground

The ball will hit the ground when h(t) = 0. So we must find the value of t such that:

$h(t)=-16t^2+80t+96=0.$

Solving this equation is equivalent to finding the roots of a second-order polynomial:

$h(t)=a*t^2+b*t+c.$

Where:

• a = -16,

,

• b = 80,

,

• c = 96.

The roots of this polynomial are given by the following formula:

s

Answer

A. The maximum height of the ball is 196 ft.

B.