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?
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To find the number of seconds after launching when the rocket returns to the ground, we need to find when the height, h(t), equals zero.
Hence, we can set the equation h(t) = 0 and solve for t:
-4.9t^2 + 25t + 12 = 0
This quadratic equation can be solved using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
The quadratic formula 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 case, a = -4.9, b = 25, and c = 12.
Plugging these values into the quadratic formula, we have:
t = (-25 ± √(25^2 - 4*(-4.9)*12)) / (2*(-4.9))
Calculating the discriminant inside the square root:
t = (-25 ± √(625 + 235.2)) / (-9.8)
Simplifying:
t = (-25 ± √(860.2)) / (-9.8)
Now, let's calculate the two possible solutions for t:
t₁ = (-25 + √(860.2)) / (-9.8) ≈ 4.55 seconds
t₂ = (-25 - √(860.2)) / (-9.8) ≈ -0.67 seconds
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the rocket will return to the ground approximately 4.55 seconds after launching.