Question

A small rocket is shot upward from the surface of the earth with an initial velocity of 120 meters per second. The position function of the rocket at time t is modeled by the function s(t) = 4.9t² + 120t. When does the rocket hit the ground?

Answer 1

To find when the rocket hits the ground using the position function s(t) = 4.9t² + 120t, it should be noted that the equation seems to have an error. The corrected position function should account for the negative acceleration due to gravity. When properly corrected, the rocket will hit the ground approximately 24.49 seconds after launch.

Step-by-step explanation:

To determine when the rocket hits the ground, we need to find when the position function s(t) equals zero. The given position function is s(t) = 4.9t² + 120t, which is a quadratic equation. Setting s(t) to zero gives us the equation to solve:

0 = 4.9t² + 120t

We can factor out a t to simplify the equation:

0 = t(4.9t + 120)

This gives us two possible solutions t=0 and t=-120/4.9. The t=0 solution represents the launch time, and the second solution will give us the time when the rocket will hit the ground after launch. Solving for that time:

t = -120/4.9 ≈ -24.49 seconds

Because time cannot be negative, we disregard the negative value, which arises due to the orientation of the quadratic function. This leaves us with only the t=0 solution, which represents the instant immediately after the rocket is launched and before it starts its ascent. In a real-world scenario, the rocket would continue on its trajectory until gravity slows it down and eventually causes it to fall back to the ground, giving us a positive time value. However, since we have a negative value, this indicates that there is a mistake in the position function. The initial velocity term should come with a negative acceleration term due to gravity, indicating that the correct position function was likely intended to be s(t) = -4.9t² + 120t.

If we correct the position function and resolve for t when s(t) = 0, we get:

0 = -4.9t² + 120t

Factoring out t again:

0 = t(-4.9t + 120)

This time we get the solutions t=0 (launch time) and t=120/4.9. Solving for the positive time:

t = 120/4.9 ≈ 24.49 seconds

The rocket will hit the ground approximately 24.49 seconds after launch once we account for the correct position function.

Answer 2

To determine when the rocket will hit the ground, the quadratic equation 4.9t² + 120t = 0 is factored and solved, yielding the time as approximately 24.49 seconds after launch, which is the only physically meaningful solution.

Step-by-step explanation:

The student is asking when a rocket, which has a position function s(t) = 4.9t² + 120t, will hit the ground after being shot upward from the surface of the earth with an initial velocity of 120 meters per second. When the rocket hits the ground, its position s(t) will be 0. To find when that happens, we need to solve the quadratic equation 4.9t² + 120t = 0 for the time t.

First, we factor out the common factor, which is t, from both terms of the equation

t(4.9t + 120) = 0

This gives us two possible solutions for when the rocket hits the ground: t = 0 seconds (which is the time of launch) and t = -120/4.9 seconds. Since a negative time doesn't make sense for this physical scenario, we ignore that solution. To solve t = -120/4.9, we do the following calculation

t = -120/4.9 ≈ -24.49

Thus, the positive time when the rocket hits the ground is approximately 24.49 seconds after launch. It is important to note that a quadratic equation can yield two solutions, but in the context of this problem, only the positive value of time is relevant since time cannot go backwards.