Question

A person standing close to the edge on top of a 96-foot building throws a ball vertically upward. The quadratic function (t) - - 161+ 804 + 96 models the ball's height about the ground, A(t), in feet, e

seconds after it was thrown.
a) What is the maximum height of the ball?
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feet
b) How many seconds does it take until the ball hits the ground?
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seconds

Answer 1

196 ft

6 seconds

Explanation:

Solution:-

We have a quadratic time dependent model of the ball trajectory which is thrown from the top of a 96-foot building as follows:

`y(t) = -16t^2 + 80t + 96`

The height of the ball is modeled by the distance y ( t ) which changes with time ( t ) following a parabolic trajectory. To determine the maximum height of the ball we will utilize the concepts from " parabolas ".

The vertex of a parabola of the form ( given below ) is defined as:

`f ( t ) = at^2 + bt + c`

Vertex:
`t = (-b)/(2a)`

- The modelling constants are: a = -16 , b = 80.

`t = (-80)/(-32) = 2.5 s`

- Now evaluate the given function " y ( t ) " for the vertex coordinate t = 2.5 s. As follows:

`y ( 2.5 ) = -16 ( 2.5 )^2 + 80*(2.5) + 96\\\\y ( 2.5 ) = 196 ft\\`

Answer: The maximum height of the ball is 196 ft at t = 2.5 seconds.

- The amount of time taken by the ball to hit the ground can be determined by solving the given quadratic function of ball's height ( y ( t ) ) for the reference ground value "0". We can express the quadratic equation as follows:

`y ( t ) = -16t^2 + 80t + 96 = 0\\\\-16t^2 + 80t + 96 = 0`

Use the quadratic formula and solve for time ( t ) as follows:

`t = (-b +/- √(b^2 - 4 ac) )/(2a) \\\\t = (-80 +/- √(80^2 - 4 (-16)(96)) )/(-32) \\\\t = (-80 +/- 112 )/(-32) = 2.5 +/- (-3.5 )\\\\t = -1, 6`

Answer: The value of t = -1 is ignored because it lies outside the domain. The ball hits the ground at time t = 6 seconds.