y = -16x^2 + 148x +128

Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground.

y=-16x^2+148x+128

Answer 1

-16t^2 + 148x + 128 = 0

16t^2 - 148x - 128 = 0

8t^2 - 74x -64 = 0

4t^2 -37x -32 = 0

t= 37/8 + (1/8)sqr(37^2 +16(32)

t = about 10.046 seconds to reach the ground

about 10 seconds

-16t^2 + 148x + 128

-16(10)^2 + 148(10) + 128 = -1600 + 1480 + 128 = -1600+ 1608 = 8

so slightly more than 10 seconds to reach the ground

in 10 seconds it's still 8 feet above ground but accelerating very fast