Hi Corrine!
If you take a look at the equation provided to represent the height in meters for the rocket as a function of time, you have a quadratic function (a polynomial with a degree of 2 will indicate it's a quadratic). I'm assuming you meant to write -4.9t2.
H(t) = -4.9t2 + 184t + 129
If you think of how a rocket would launch and then ultimately fall into the ocean water, it will form a U-shaped curve, just like a quadratic equation will give us a U-shaped curve, a parabola.
The parabola will be concaving downward as indicated by the negative coefficient of the quadratic term (-4.9).
Knowing all of this, the rocket splashing down to the water (sea level) is equivalent to when the parabola hits the x-axis, and where we hit the x-axis (used to measure time t in this case) is when the height (used for the y-axis) is equal to 0.
Therefore to solve, we will set H(t) = 0.
-4.9t2+184t+129 = 0 with
a = -4.9
b = 184
c = 129
If we use the quadratic formula to solve (t = (-b + - sqrt(b2 - 4ac))/2a), we will get t = -0.688 or 38.239 approximately. Of course, time cannot be expressed as a negative value here, so the answer is 38.239 seconds, which would be the amount of time it takes for the rocket to splash into the water.
For the peak height, this is where our vertex of the parabola will be located before the rocket begins to fall. To find the vertex, we can look for the axis of symmetry first to help us find the x-coordinate (t) of the vertex simultaneously, which is expressed by t = -b/2a.
t = -184/[2(-4.9)] = approximately 18.77 seconds in is when the rocket will reach its peak (do not round until the very end to avoid potential rounding errors). If we plug this in for t in the original quadratic function H(t):
H(18.77) = -4.9(18.77)2 + 184(18.77) + 129, we will get approximately 1,856.35 meters as the peak height.
Check to see if the question asks you to round to a different decimal place, otherwise that is how you evaluate these two problems. Let me know if this helps.
Thanks!
