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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

User Emeh
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1 Answer

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Step-by-step 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=(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.

User Cesarina
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