solving projectile motion. an object is launched straight into the air. the projectile motion of the object…

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?

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?

Answer

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.