MATH SOLVE

4 months ago

Q:
# A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function: f(t) = -16t^2 + 16t + 32 Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?A: 0.5 < t < 1B: 0.5 < t < 2C: 1 < t < 2 D: 1 < t < 1.5

Accepted Solution

A:

To solve this we are going to find the maximum eight of the parabola first. Since the parabola is opening downwards, the maximum height of the ball will be the t-coordinate of its vertex.

Remember that to find the vertex [tex](h,k)[/tex] of a parabola of the form [tex]a x^{2} +bx+c[/tex] we use the vertex formula: [tex]h= \frac{-b}{2a} [/tex], and then, we evaluate our function at [tex]h[/tex] to find [tex]k[/tex].

We can infer for our parabola that [tex]a=-16[/tex] and [tex]b=16[/tex], so lets replace those values in our formula:

[tex]h= \frac{-b}{2a} [/tex]

[tex]h= \frac{-16}{2(-16)} [/tex]

[tex]h= \frac{-16}{-32} [/tex]

[tex]h=0.5[/tex]

Now we now that the ball reaches its maximum height in 0.5 seconds.

Next, we are going to determine how long it takes for the ball to hit the ground. To do that we are going to set the time [tex]f(t)[/tex] equal to zero and solve for [tex]t[/tex]:

[tex]0=-16t^2+16t+32[/tex]

[tex]16t^2-16t-32=0[/tex]

[tex]16(t^2-t-2)=0[/tex]

[tex]t^2-t-2=0[/tex]

[tex](t+1)(t-2)=0[/tex]

[tex]t=-1[/tex] or [tex]t=2[/tex]

Since time cannot be negative, the answer is [tex]t=2[/tex]

We can conclude that a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground is B: 0.5 < t < 2

Remember that to find the vertex [tex](h,k)[/tex] of a parabola of the form [tex]a x^{2} +bx+c[/tex] we use the vertex formula: [tex]h= \frac{-b}{2a} [/tex], and then, we evaluate our function at [tex]h[/tex] to find [tex]k[/tex].

We can infer for our parabola that [tex]a=-16[/tex] and [tex]b=16[/tex], so lets replace those values in our formula:

[tex]h= \frac{-b}{2a} [/tex]

[tex]h= \frac{-16}{2(-16)} [/tex]

[tex]h= \frac{-16}{-32} [/tex]

[tex]h=0.5[/tex]

Now we now that the ball reaches its maximum height in 0.5 seconds.

Next, we are going to determine how long it takes for the ball to hit the ground. To do that we are going to set the time [tex]f(t)[/tex] equal to zero and solve for [tex]t[/tex]:

[tex]0=-16t^2+16t+32[/tex]

[tex]16t^2-16t-32=0[/tex]

[tex]16(t^2-t-2)=0[/tex]

[tex]t^2-t-2=0[/tex]

[tex](t+1)(t-2)=0[/tex]

[tex]t=-1[/tex] or [tex]t=2[/tex]

Since time cannot be negative, the answer is [tex]t=2[/tex]

We can conclude that a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground is B: 0.5 < t < 2