Question

an object is launched into the air. the projectile motion of the object can be modeled using the function ( h(t)=-16t^{2}+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 the initial height

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

Step2: Analyze the initial velocity

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

Step3: Find when the object hits the ground

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

Step4: Find the height at \(t = 3\)

Substitute \(t = 3\) into \(h(t)\): \(h(3)=-16\times3^{2}+72\times3+5=-16\times9+216 + 5=-144+216+5=77
eq173\).

Step5: Check \(h(0)\)

As calculated in Step1, \(h(0)=5
eq0\).

Answer:

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