Question

an arrow is launched upward with a velocity of 192 feet per second from the top of a 35 - foot building. what is the maximum height attained by the arrow?

Explanation:

Step1: Write the height function

The height function for vertical - motion is \(h(t)=-16t^{2}+v_{0}t + h_{0}\), where \(v_{0}\) is the initial velocity and \(h_{0}\) is the initial height. Given \(v_{0}=192\) and \(h_{0}=35\), so \(h(t)=-16t^{2}+192t + 35\).

Step2: Find the time \(t\) at which the maximum height occurs

For a quadratic function \(y = ax^{2}+bx + c\) (\(a
eq0\)), the \(x\) - coordinate of the vertex (in our case, \(t\) for the function \(h(t)\)) is given by \(t=-\frac{b}{2a}\). Here, \(a=-16\) and \(b = 192\). Then \(t=-\frac{192}{2\times(-16)}=\frac{-192}{-32}=6\).

Step3: Find the maximum height

Substitute \(t = 6\) into the height function \(h(t)\). \(h(6)=-16\times6^{2}+192\times6 + 35\). First, calculate \(-16\times6^{2}=-16\times36=-576\), \(192\times6 = 1152\). Then \(h(6)=-576+1152 + 35=611\).

Answer:

611