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+266x+102 The amount of time it takes for the rocket to hit the ground is ____ seconds

Answer 1

Answer:

17 seconds

Step-by-step explanation:

Solution:-

The trajectory of a rocket launched is modeled by a parabolic path via a quadratic equation as follows:

                               $y = -16x^2 + 266x +102$

Where,

                   y: The height of the rocket from ground

                   x: The time taken since rocket launch

We are to determine the time ( x ) it takes for the rocket to hit the ground. This can be determined by setting the quantity ( y ) to zero as follows

                             $y = -16x^2 + 266x +102 = 0\\\\ -16x^2 + 266x +102 = 0$

solve the quadratic equation using quadratic formula as follows:

                             $x = \frac{-b +/- \sqrt{b^2 - 4ac} }{2a} \\\\x = \frac{-266 +/- \sqrt{266^2 - 4*(-16)*(102)} }{2*(-16)} \\\\x = \frac{-266 +/- 278 }{-32} \\\\x = 8.3125 +/- 8.6875\\\\x = 17 , -0.375$

Answer: Ignore the negative value for time " -0.375 ". Hence, the rocket hits the ground at time t = 17 seconds