Question

solving projectile motion. an object is launched straight into the air. the projectile motion of the object can be modeled using $h(t)=96t - 16t^{2}$, where $t$ is the time since launch and $h(t)$ is the height in feet of the projectile after time $t$ in seconds. when will the object be 144 feet in the air? when will the object hit the ground?

Explanation:

Step1: Set up equation for height of 144 feet

Set $h(t)=144$, so $144 = 96t-16t^{2}$. Rearrange to $16t^{2}-96t + 144=0$. Divide by 16 to get $t^{2}-6t + 9 = 0$.

Step2: Solve the quadratic equation

Factor the quadratic: $(t - 3)^{2}=0$. Solving for $t$ gives $t = 3$.

Step3: Set up equation for hitting the ground

When the object hits the ground, $h(t)=0$. So $0=96t-16t^{2}$. Factor out $16t$: $16t(6 - t)=0$.

Step4: Solve for hitting - ground time

Set each factor equal to zero. $16t=0$ gives $t = 0$ (corresponds to launch time) and $6 - t=0$ gives $t = 6$.

Answer:

The object is 144 feet in the air at $t = 3$ seconds. The object hits the ground at $t = 6$ seconds.