The company will be launching a firework from the ground, but you want to increase the time it stays in the air by 2 seconds to build up the suspense. How can the company make this happen without changing the length of the fuse? How will this affect the maximum height of the firework? g(t) = -16t2 + 224t + 120

find the answer in the explanation.

Step-by-step explanation:

To increase the time the fireworks stays in the air by 2 seconds to build up the suspense, 2 should be added to the time t.

T + 2 = t

T = t - 2

Substitute T for t in the function g(t)

The company will make this happen without changing the length of the fuse by substituting T for t which will lead to

g(t) = -16(t - 2)^2 + 224(t - 2) + 120

How will this affect the maximum height of the firework?

Initially the function is:

g(t) = -16t2 + 224t + 120

Where the time at maximum height is found by using the formula

t = -b/2a

b = 224, a = -16

Substitutes both into the formula

t = -224/2(-16)

t = -224/-32

t = 7

Substitute 7 for t in the first equation

g(t) = -16(7)^2 + 224(7) + 120

g(t) = -16(49) + 224(7) + 120

g(t) = -784 + 1568 + 120

g(t) = 904 metres

But when the time is delayed by 2 seconds in the air, the maximum height will be

g(t) = -16( 7 - 2 )^2 + 224(7 - 2) + 120

g(t) = -16(5)^5 + 224(5) + 120

g(t) = -16(25) + 224(5) + 120

g(t) = -400 + 1120 + 120

g(t) = 840

The effect will surely reduce the maximum height of the fireworks.

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: