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

17 seconds

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 = (-b +/- √(b^2 - 4ac) )/(2a) \\\\x = (-266 +/- √(266^2 - 4*(-16)*(102)) )/(2*(-16)) \\\\x = (-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