Answer:
y = -16 (x - 1)^2 + 112
The object lands on the ground in approximately 3.6s
Explanation:
The equation given is that of a parabola.
Now the maximum (local) point of a parabola is the vertex. Therefore, if we want to rewrite our function in the form that would be used to find the maximum height, then that form must be the vertex form of a parabola.
The vertex form of a parabola is
where (h, k) is the vertex.
The only question is, what is the vertex for our function h(t)?
Remember that if we have an equation of the form
then the x-coordinate of the vertex is
Now in our case b = 32 and a = -16; therefore,
We've found the value of the x-coordinate of the vertex. What about the y-coordinate? To get the y-coordinate, we put x = 1 into h(t) and get
Hence, the y-coordindate is k = 112.
Therefore, the vertex of the parabola is (1, 112).
With the coordinates of the vertex in hand, we now write the equation of the parabola in vertex form.
The only problem is that we don't know what the value of a is. How do we find a?
Note that the point (0, 96) lies on the parabola. In other words,
Therefore, the vertex form of the parabola must also contain the point (0, 96).
Putting in t = 0, h = 96 into the vertex form gives
subtracting 112 from both sides gives
With the value of a in hand, we can finally write the equation of the parabola on vertex form.
Now when does the object hit the ground? In other words, for what value of t is h(t) = 0? To find out we just have to solve the following for t.
We could either use h(t) = -16t^2 + 32t + 96 or the h(t) = -16(t - 1)^2 + 112 for the above equation. But it turns out, the vertex form is more convenient.
Thus we solve,
Now subtracting 112 from both sides gives
Dividing both sides by -16 gives
taking the square root of both sides gives
adding 1 to both sides gives
Hence, the two solutions we get are
Now since time cannot take a negative value, we discard the second solution and say that t = 3.6 is our valid solution.
Therefore, it takes about 3.6 seconds for the object to hit the ground.
