Respuesta :
Answer:
196 ft
6 seconds
Step-by-step 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:
[tex]y(t) = -16t^2 + 80t + 96[/tex]
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:
[tex]f ( t ) = at^2 + bt + c[/tex]
Vertex: [tex]t = \frac{-b}{2a}[/tex]
- The modelling constants are: a = -16 , b = 80.
[tex]t = \frac{-80}{-32} = 2.5 s[/tex]
- Now evaluate the given function " y ( t ) " for the vertex coordinate t = 2.5 s. As follows:
[tex]y ( 2.5 ) = -16 ( 2.5 )^2 + 80*(2.5) + 96\\\\y ( 2.5 ) = 196 ft\\[/tex]
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:
[tex]y ( t ) = -16t^2 + 80t + 96 = 0\\\\-16t^2 + 80t + 96 = 0[/tex]
Use the quadratic formula and solve for time ( t ) as follows:
[tex]t = \frac{-b +/- \sqrt{b^2 - 4 ac} }{2a} \\\\t = \frac{-80 +/- \sqrt{80^2 - 4 (-16)(96)} }{-32} \\\\t = \frac{-80 +/- 112 }{-32} = 2.5 +/- (-3.5 )\\\\t = -1, 6[/tex]
Answer: The value of t = -1 is ignored because it lies outside the domain. The ball hits the ground at time t = 6 seconds.
