Question

modeling cr- quadratics
the path of a projectile can be modeled by the function $h(t)=-16t^{2}+v_{0}t + h_{0}$, where h(t) is the height of the projectile (in feet), t is the time (in seconds), $v_{0}$ is the initial velocity (in feet per second), and $h_{0}$ is the initial height (in feet).
during world war i, mortars were fired from trenches 3 feet below the ground level. the mortars had an initial velocity of 150 feet per second.

• create a model to represent the relationship between the height of the mortar, h(t), at any given time, t. explain how you determined your model.
• use your model to determine at what time the mortar will hit the ground (rounded to the nearest tenth of a second). show your work and/or explain your reasoning.

Explanation:

Step1: Determine model coefficients

Given $v_0 = 150$ (initial - velocity) and $h_0=- 3$ (3 feet below ground level), the model is $h(t)=-16t^{2}+150t - 3$.

Step2: Set $h(t) = 0$ to find ground - hitting time

We have the quadratic equation $-16t^{2}+150t - 3=0$. For a quadratic equation $ax^{2}+bx + c = 0$ ($a=-16$, $b = 150$, $c=-3$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

Step3: Calculate the discriminant

$\Delta=b^{2}-4ac=(150)^{2}-4\times(-16)\times(-3)=22500 - 192 = 22308$.

Step4: Find the values of $t$

$t=\frac{-150\pm\sqrt{22308}}{-32}=\frac{-150\pm149.36}{-32}$.
We get two solutions for $t$:
$t_1=\frac{-150 + 149.36}{-32}=\frac{-0.64}{-32}=0.02$ and $t_2=\frac{-150 - 149.36}{-32}=\frac{-299.36}{-32}\approx9.4$.
We discard $t = 0.02$ (corresponds to the initial moment when it is fired), so the mortar hits the ground at $t\approx9.4$ seconds.

Answer:

The model is $h(t)=-16t^{2}+150t - 3$. The mortar hits the ground at approximately $9.4$ seconds.