Question

A rocket is launched with an initial upward velocity of 320 feet per second from an initial height of 15 feet.

The function h (t) = -1672 + 320t + 15 models the height of the rocket, in feet, t seconds after it was
launched. For how many seconds is the height of the rocket greater than or equal to 1.039 feet?

Answer 1

The quadratic function h(t) needs to be set with the inequality h(t) ≥ 1.039 and solved to determine the time intervals for which the height of the rocket is at least 1.039 feet. Solving the quadratic will give two time values which represent the interval when the rocket's height is greater than or equal to 1.039 feet.

Step-by-step explanation:

The function h(t) = -16t2 + 320t + 15 represents the height of the rocket in feet, t seconds after launch. To find for how many seconds the height of the rocket is greater than or equal to 1.039 feet, we need to determine the time intervals when h(t) ≥ 1.039.

We set the inequality h(t) ≥ 1.039 and solve for t:

1. First, subtract 1.039 from both sides to get a quadratic equation: -16t2 + 320t + 13.961 ≥ 0.
2. Factor the quadratic or use the quadratic formula to find the roots of the equation.
3. The result will give us two values of t, say t1 and t2, and the rocket's height is greater than or equal to 1.039 feet for t in the interval [t1, t2].
4. Determine the positive root since time cannot be negative, and this will be the time interval we are looking for.