The height of a rocket is launched from a platform 12 meters above the ground is modeled by the function h(t)=-4.9t^2+25t+12 where h(t) represents the height of the rocket in meters and t represents the number of seconds after launching. How many seconds after launching will the rocket return to the ground?
h(t)=0
so
4.9t^2-25t-12=0
use the quadratic equation, solve for t.
To find the time it takes for the rocket to return to the ground, we need to determine when the height of the rocket is equal to zero.
In this case, the height function is given as h(t) = -4.9t^2 + 25t + 12, where h(t) represents the height of the rocket in meters and t represents the time in seconds.
To find when the height is equal to zero, we can set h(t) = 0 and solve for t:
-4.9t^2 + 25t + 12 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -4.9, b = 25, and c = 12. Substituting these values into the formula, we get:
t = (-25 ± √(25^2 - 4(-4.9)(12))) / (2(-4.9))
Simplifying further:
t = (-25 ± √(625 + 235.2)) / (-9.8)
t = (-25 ± √860.2) / (-9.8)
The two possible values for t are obtained by using both "+" and "−" in the formula above. Let's calculate these two values separately:
For t = (-25 + √860.2) / (-9.8):
t ≈ 5.71 seconds
For t = (-25 - √860.2) / (-9.8):
t ≈ -0.28 seconds
Since time cannot be negative in this context, the rocket will return to the ground approximately 5.71 seconds after launching.