Question

quadratic modeling - independent practice
directions: for problems 1 - 2, write a vertical motion model in the form h(t)=-16t² + v₀t+ h₀. then find (a) the time to maximum height, and (b) the maximum height. for problem 3, write a quadratic function to model the situation.

1. initial velocity v₀ = 32 ft/s; initial height h₀ = 20 ft.

• equation h(t)=
• t_max=
• h_max=

1. initial velocity v₀ = 120 ft/s; initial height h₀ = 50 ft.

• equation h(t)=
• t_max=
• h_max=

1. a rectangular patio has length four times its width w. a 3 - ft brick border surrounds it. write a quadratic function for the total area (patio + border) in terms of w.

• a_total(w)=

(optional) also find patio - only area and border - only area.
helpful formulas:

• h(t)=-16t² + v₀t + h₀
• vertex time: t_max = v₀/32
• maximum height: h(t_max)

Explanation:

Step1: Write the vertical - motion equation for problem 1

Given $v_0 = 32$ ft/s and $h_0=20$ ft, substitute into $h(t)=-16t^{2}+v_0t + h_0$.
$h(t)=-16t^{2}+32t + 20$

Step2: Find the time to maximum height for problem 1

Use the formula $t_{max}=\frac{v_0}{32}$. Substitute $v_0 = 32$ ft/s.
$t_{max}=\frac{32}{32}=1$ s

Step3: Find the maximum height for problem 1

Substitute $t = 1$ into $h(t)=-16t^{2}+32t + 20$.
$h(1)=-16\times1^{2}+32\times1 + 20=-16 + 32+20=36$ ft

Step4: Write the vertical - motion equation for problem 2

Given $v_0 = 120$ ft/s and $h_0 = 50$ ft, substitute into $h(t)=-16t^{2}+v_0t + h_0$.
$h(t)=-16t^{2}+120t + 50$

Step5: Find the time to maximum height for problem 2

Use the formula $t_{max}=\frac{v_0}{32}$. Substitute $v_0 = 120$ ft/s.
$t_{max}=\frac{120}{32}=\frac{15}{4}=3.75$ s

Step6: Find the maximum height for problem 2

Substitute $t = 3.75$ into $h(t)=-16t^{2}+120t + 50$.
$h(3.75)=-16\times(3.75)^{2}+120\times3.75 + 50=-16\times14.0625+450 + 50=-225+450 + 50=275$ ft

Step7: Write the quadratic function for the total area in problem 3

The width of the patio is $w$, the length of the patio is $l = 4w$. The width of the patio + border is $w + 6$ and the length of the patio+border is $4w+6$.
$A_{total}(w)=(4w + 6)(w + 6)=4w^{2}+24w+6w + 36=4w^{2}+30w + 36$
The patio - only area $A_{patio}=4w\times w=4w^{2}$.
The border - only area $A_{border}=A_{total}-A_{patio}=4w^{2}+30w + 36-4w^{2}=30w + 36$

Answer:

1.

• Equation $h(t)=-16t^{2}+32t + 20$
• $t_{max}=1$ s
• $h_{max}=36$ ft

2.

• Equation $h(t)=-16t^{2}+120t + 50$
• $t_{max}=3.75$ s
• $h_{max}=275$ ft

3.

• $A_{total}(w)=4w^{2}+30w + 36$
• Patio - only area: $4w^{2}$
• Border - only area: $30w + 36$