Question

a projectile is shot upward from the top of a 112 - foot building at an initial upward velocity of 96 feet per second. the function ( h(t)=-16 t^{2}+96 t + 112 ) describes the height of the projectile in feet as a function of time in seconds. how many seconds from launch will it take for the projectile to fall to the ground? 7 seconds 5 seconds 3 seconds 1 seconds

Explanation:

Step1: Set the height function equal to 0

When the projectile falls to the ground, \(h(t)=0\). So we have the equation \(-16t^{2}+96t + 112=0\).
Divide through by \(-16\) to simplify: \(t^{2}-6t - 7=0\).

Step2: Factor the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-6\), \(c=-7\)), we use the formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor.
We factor \(t^{2}-6t - 7=(t - 7)(t+ 1)=0\).

Step3: Solve for \(t\)

Set each factor equal to zero:
\(t - 7=0\) gives \(t = 7\); \(t+1=0\) gives \(t=-1\).
Since time \(t\geq0\), we discard \(t=-1\).

Answer:

7 seconds