A model rocket is launched with an initial velocity of 250 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after launch will the rocket be 730 ft above the ground? Round to the nearest hundredth of a second.
solve
730 = -16t^2 + 250t
16t^2 - 250t + 730 = 0
(use the quadratic equation formula)
To find how many seconds after launch the rocket will be 730 ft above the ground, we need to solve the equation h = 730 for t. In this case, the equation is:
-16t^2 + 250t = 730
To solve this quadratic equation, we can rearrange it to:
16t^2 - 250t + 730 = 0
Next, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 16, b = -250, and c = 730. Substituting these values into the quadratic formula, we have:
t = (-(-250) ± √((-250)^2 - 4 * 16 * 730)) / (2 * 16)
Simplifying further:
t = (250 ± √(62500 - 46720)) / 32
t = (250 ± √15780) / 32
Now, we have two possible solutions for t:
t1 = (250 + √15780) / 32
t2 = (250 - √15780) / 32
Calculating these two values:
t1 ≈ 13.07 seconds
t2 ≈ 12.68 seconds
Therefore, the rocket will be 730 ft above the ground approximately 13.07 seconds and 12.68 seconds after launch.