Question

analyzing projectile motion problems
an object is launched into the air. the projectile motion of the object can be modeled using the function h(t)=-16t² + 72t + 5, where t is the time in seconds since the launch and h(t) represents the height in feet of the object after t seconds. what is true about the projectile motion of this object? check all that apply.
the initial height is 5 feet.
the initial velocity of the object is -72 feet/second.
the object will hit the ground after approximately 4.57 seconds.
after 3 seconds, the object is 173 feet high.
at t = 0, h(t)=0.

Explanation:

Step1: Find initial height

When $t = 0$, $h(0)=-16\times0^{2}+72\times0 + 5=5$. So the initial height is 5 feet.

Step2: Analyze initial velocity

The general form of a projectile - motion height - function is $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $v_{0}$ is the initial velocity and $h_{0}$ is the initial height. Comparing with $h(t)=-16t^{2}+72t + 5$, the initial velocity $v_{0}=72$ feet/second, not - 72 feet/second.

Step3: Find when the object hits the ground

Set $h(t)=0$, so $-16t^{2}+72t + 5 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a=-16$, $b = 72$, $c = 5$. Then $t=\frac{-72\pm\sqrt{72^{2}-4\times(-16)\times5}}{2\times(-16)}=\frac{-72\pm\sqrt{5184 + 320}}{-32}=\frac{-72\pm\sqrt{5504}}{-32}=\frac{-72\pm74.2}{-32}$. We take the positive root $t=\frac{-72 + 74.2}{-32}\approx4.57$ seconds.

Step4: Find height at $t = 3$

When $t = 3$, $h(3)=-16\times3^{2}+72\times3+5=-16\times9 + 216+5=-144 + 216+5=77$ feet, not 173 feet.

Step5: Evaluate $h(t)$ at $t = 0$

We already found that when $t = 0$, $h(0)=5
eq0$.

Answer:

The initial height is 5 feet.
The object will hit the ground after approximately 4.57 seconds.