Q:

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?

Accepted Solution

A:
Answer:h(x) = -16x² + 192x + 208 784ft6 sec13 secStep-by-step explanation:a)h(x) = -16x² +vx + h[tex]_{o}[/tex]here v represent velocity          [tex]h_{o}[/tex] represent initial height of launch        h(x) = -16x² + 192x + 208b)h(x) = -16x² + 192x + 208here a = -16         b = 192         c = 208x = -b/2a   = -192/2(-16)   = 6plug this value in the equationh(x) = -16(6)² + 192(6) + 208       = 784fte)Plug h(x)=0 in the equation0 = -16x² + 192x + 208divide equation by -16x² - 12x - 13 = 0 Factors1x * -13x = -131x - 13x = -12Factorised formx² - 12x - 13 = 0 x² + x - 13x - 13 = 0 x(x+1) -13(x+1) = 0 (x+1)(x-13) = 0x = -1x = 13Since time can not be negative so we will reject x = -1