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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(t) = - 16t^2 + 116t + 96 models the balls height about the ground, h(t), in feet, t seconds after it was thrown.

a) What is the maximum height of the ball? __ feet

b) How many seconds does it take until the ball hits the ground? __ seconds

User Marek Tihkan
by
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1 Answer

26 votes
26 votes

Answer:

the answer is 306.25 feet and b is 8 sec

Explanation:

Explanation

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

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:

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

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

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:

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

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.

User Shantanu Pathak
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