Mathematics, 26.07.2020 23:01 harveyangel123p2tjae

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

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Answer from: pevin120

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:

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: $t = \frac{-b}{2a}$

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

$t = \frac{-80}{-32} = 2.5 s$

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

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

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:

$t = \frac{-b +/- \sqrt{b^2 - 4 ac} }{2a} \\\\t = \frac{-80 +/- \sqrt{80^2 - 4 (-16)(96)} }{-32} \\\\t = \frac{-80 +/- 112 }{-32} = 2.5 +/- (-3.5 )\\\\t = -1, 6$

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

Answer from: Quest

answer: -1

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

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