Question

a boomerang is thrown in the air and it follows a path described by the equation f(t)= - 16t² + 192t + 5. height in feet and t is the time in seconds. the initial height of the boomerang is feet. the boomerang will hit the ground between seconds and seconds. the maximum height of the boomerang is feet after about seconds.

Explanation:

Step1: Identify the initial - height

The initial height is when \(t = 0\). Substitute \(t=0\) into the height - function \(h(t)=-16t^{2}+192t + 5\).
\[h(0)=-16\times0^{2}+192\times0 + 5=5\]

Step2: Find the time of maximum height

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex (in our case, the time \(t\) of the maximum height) is given by \(t=-\frac{b}{2a}\). Here, \(a=-16\) and \(b = 192\).
\[t=-\frac{192}{2\times(-16)}=\frac{-192}{-32}=6\]
Then substitute \(t = 6\) into the height - function:
\[h(6)=-16\times6^{2}+192\times6 + 5=-16\times36+1152 + 5=-576+1152 + 5=581\]

Step3: Find the time when it hits the ground

Set \(h(t)=0\), so we have the quadratic equation \(-16t^{2}+192t + 5=0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a=-16\), \(b = 192\), and \(c = 5\).
\[t=\frac{-192\pm\sqrt{192^{2}-4\times(-16)\times5}}{2\times(-16)}=\frac{-192\pm\sqrt{36864 + 320}}{-32}=\frac{-192\pm\sqrt{37184}}{-32}=\frac{-192\pm192.83}{-32}\]
We take the positive root \(t=\frac{-192 + 192.83}{-32}\approx12.03\) (we ignore the negative root since time cannot be negative).

Answer:

The initial height of the boomerang is 5 feet. The maximum height of the boomerang is 581 feet after about 6 seconds. The boomerang will hit the ground between 12 seconds and 13 seconds.