an object is launched into the air. the projectile motion of the object can be modeled using the function…

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.

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.

Answer

Explanation:

Step1: Find initial height

The initial height is when $t = 0$. Substitute $t=0$ into $h(t)=-16t^{2}+72t + 5$. So $h(0)=-16\times0^{2}+72\times0 + 5=5$. The initial height is 5 feet.

Step2: Find 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.

Step3: Find when the object hits the ground

The object hits the ground when $h(t)=0$. So we solve the quadratic equation $-16t^{2}+72t + 5 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a=-16$, $b = 72$, and $c = 5$. 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 $t = 3$

Substitute $t = 3$ into $h(t)=-16t^{2}+72t + 5$. Then $h(3)=-16\times3^{2}+72\times3+5=-16\times9 + 216+5=-144 + 216+5=77$ feet.

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

We already found in Step 1 that $h(0)=5\neq0$.

Answer:

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