Question

if an object is projected upward from ground level with an initial velocity of 80 ft per sec, then its height in feet after t seconds is given by ( s(t)=-16 t^{2}+80 t ). find the number of seconds it will take to reach its maximum height. what is this maximum height? the object will take ( square ) second(s) to reach its maximum height. (simplify your answer.) the maximum height reached by the object is ( square ) feet. (simplify your answer.)

Explanation:

Step1: Find the time to reach maximum height

For a quadratic function \(s(t)=at^{2}+bt + c\) (here \(a=-16\), \(b = 80\), \(c = 0\)), the time \(t\) at which the vertex (maximum for \(a<0\)) occurs is given by the formula \(t=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 80\) into the formula:
\(t=-\frac{80}{2\times(-16)}=\frac{-80}{-32}=\frac{80}{32}=\frac{5}{2}=2.5\)

Step2: Find the maximum height

Substitute \(t = 2.5\) into the function \(s(t)=-16t^{2}+80t\).
\(s(2.5)=-16\times(2.5)^{2}+80\times(2.5)\)
First, calculate \((2.5)^{2}=6.25\)
\(s(2.5)=-16\times6.25 + 200\)
\(-16\times6.25=-100\)
\(s(2.5)=-100 + 200=100\)

Answer:

The object will take \(2.5\) second(s) to reach its maximum height.
The maximum height reached by the object is \(100\) feet.