Question

An object is thrown upward from the top of a 48-foot building with an initial velocity of 32 feetquadratic equation h = - 16t^2+ 32t + 48. When will the object hit the ground?

Answer 1

SOLUTION

The object would hit the ground at the root of the equation, that is where h = 0

So we have

$\begin{gathered} h=-16t^2+32t+48=0 \\ -16t^2+32t+48=0 \\ 16t^2-32t-48=0 \\ divide\text{ through by 16} \\ t^2-2t-3=0 \end{gathered}$

Solving the quadratic equation, we have

$\begin{gathered} t^2-3t+t-3=0 \\ t(t-3)+1(t-3)=0 \\ (t-3)(t+1)=0 \\ t=3,\text{ or} \\ t=-1 \end{gathered}$

So we ignore -1 as time cannot be negative.

Hence the object hit the ground in 3 seconds