Question

An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reached by the rocket?

Answer 1

Answer:

h(x) = -16x² + 192x + 208

784ft

6 sec

13 sec

Step-by-step explanation:

a)

h(x) = -16x² +vx + h$_{o}$

here v represent velocity

         $h_{o}$ represent initial height of launch

       

h(x) = -16x² + 192x + 208

b)

h(x) = -16x² + 192x + 208

here a = -16

        b = 192

        c = 208

x = -b/2a

  = -192/2(-16)

  = 6

plug this value in the equation

h(x) = -16(6)² + 192(6) + 208

      = 784ft

e)

Plug h(x)=0 in the equation

0 = -16x² + 192x + 208

divide equation by -16

x² - 12x - 13 = 0

Factors

1x * -13x = -13

1x - 13x = -12

Factorised form

x² - 12x - 13 = 0

x² + x - 13x - 13 = 0

x(x+1) -13(x+1) = 0

(x+1)(x-13) = 0

x = -1

x = 13

Since time can not be negative so we will reject x = -1