Question

Jeremy shoots a water-powered rocket from ground level with an initial vertical velocity of 96 ft. per second. The height of the object H, in feet, as a function of time, t, in seconds, is given by the function below:

H(t) = 96t - 16t²

After how many seconds is the rocket at or above 80 feet, relative to ground level?"

You can solve this problem by setting H(t) equal to 80 feet and solving for t.

Answer 1

The rocket will be at or above 80 feet for 2.5 seconds and 2 seconds.

Step-by-step explanation:

To determine the number of seconds the rocket is at or above 80 feet, we need to set the height function equal to 80 and solve for t. The height function, H(t), is given by H(t) = 96t - 16t². So, 96t - 16t² = 80. Rearranging the equation, we get 16t² - 96t + 80 = 0.

This is a quadratic equation, so we can solve it by factoring or using the quadratic formula. Factoring this equation gives us (4t - 10)(4t - 8) = 0. Setting each factor equal to 0 gives us t = 2.5 or t = 2. In this context, negative time doesn't make sense, so we only consider the positive values of t.

Therefore, the rocket will be at or above 80 feet for 2.5 seconds and 2 seconds.