Question

write a vertical motion model in the form h(t)=-16t² + v₀t + h₀. determine the maximum height of the situation. the initial velocity is 32 ft/s and the initial height is 20 ft.

Explanation:

Step1: Write the motion - model equation

Given $v_0 = 32$ ft/s and $h_0=20$ ft, the vertical - motion model is $h(t)=-16t^{2}+32t + 20$.

Step2: Find the time $t$ at which the maximum occurs

For a quadratic function $y = ax^{2}+bx + c$ ($a
eq0$), the $x$ - coordinate of the vertex (in our case, the time $t$ at which the maximum of $h(t)$ occurs) is given by $t=-\frac{b}{2a}$. Here, $a=-16$ and $b = 32$. So, $t=-\frac{32}{2\times(-16)}=\frac{-32}{-32}=1$ s.

Step3: Calculate the maximum height

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

Answer:

36 ft