Question

emma built a model rocket as a summer project. after she launches the rocket, it soars up into the sky. it runs out of fuel at a height of 450 feet when it is traveling upward with a velocity of 100 feet per second. the rocket continues upward for a time and then falls back down to the ground. which equation can you use to find how many seconds it takes the rocket to hit the ground after running out of fuel? if an object travels upward at a velocity of v feet per second from s feet above the ground, the objects height in feet, h, after t seconds can be modeled by the formula h = -16t² + vt + s. 0 = -16t² + 450t + 100 0 = -16t² + 100t + 450 to the nearest tenth of a second, how long does it take the rocket to hit the ground after running out of fuel? seconds

Explanation:

Step1: Identify variables

Given: v=100 ft/s (initial velocity), s=450 ft (initial height). Ground hit when h=0.

Step2: Substitute into formula

h=-16t²+vt+s → 0=-16t²+100t+450.

Step3: Solve quadratic equation

a=-16, b=100, c=450. Discriminant: 100²-4*(-16)(450)=38800. √38800≈196.977.
t=(-100-196.977)/(2*(-16))≈9.3 (positive solution).

Answer:

0 = -16t² + 100t + 450, 9.3