Question

if a projectile is fired straight upward from the ground with an initial speed of 128 feet per second, then its height h in feet after t seconds is given by the function h(t). the maximum height of the projectile is
(simplify your answer.)

Explanation:

1. Recall the kinematic - equation for vertical motion:

• The height function of an object in vertical - motion under the influence of gravity is given by \(h(t)=-16t^{2}+v_{0}t + h_{0}\), where \(v_{0}\) is the initial velocity and \(h_{0}\) is the initial height. In this case, the object is fired from the ground, so \(h_{0} = 0\), and the initial velocity \(v_{0}=128\) feet per second. So, \(h(t)=-16t^{2}+128t\).

1. Find the time \(t\) at which the object reaches its maximum height:

• For a quadratic function of the form \(y = ax^{2}+bx + c\) (in our case, \(a=-16\), \(b = 128\), and \(c = 0\)), the \(x\) - coordinate (in our case, the time \(t\)) of the vertex of the parabola is given by \(t=-\frac{b}{2a}\).
• Substitute \(a=-16\) and \(b = 128\) into the formula \(t=-\frac{b}{2a}\).
• \(t=-\frac{128}{2\times(-16)}=\frac{-128}{-32}=4\) seconds.

1. Find the maximum height:

• Substitute \(t = 4\) into the height function \(h(t)=-16t^{2}+128t\).
• \(h(4)=-16\times4^{2}+128\times4\).
• First, calculate \(-16\times4^{2}=-16\times16=-256\).
• Then, calculate \(128\times4 = 512\).
• \(h(4)=-256 + 512=256\) feet.

Answer:

256 feet