So this question is asking for the zeros, or the x-intercepts, of this equation. For this, we have to set h(t) to zero to solve for it.
For this, we will be using the quadratic formula, which is [tex] x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a} [/tex] (a = t^2 coefficient, b = t coefficient, and c = constant). In this case, our equation is [tex] t=\frac{-0+\sqrt{0^2-4*(-16)*144}}{2*(-16)},\frac{-0-\sqrt{0^2-4*(-16)*144}}{2*(-16)} [/tex] . From here we can solve for t.
- Firstly, solve the exponents and multiplications: [tex] t=\frac{-0+\sqrt{0+9216}}{-32},\frac{-0-\sqrt{0+9216}}{-32} [/tex]
- Next, solve the additions and subtractions: [tex] t=\frac{\sqrt{9216}}{-32},\frac{-\sqrt{9216}}{-32} [/tex]
- Next, square root 9216: [tex] t=\frac{96}{-32},\frac{-96}{-32} [/tex]
- Lastly, divide and your answers are going to be [tex] t=-3,3 [/tex]
Since we cannot have negative time, the object hits the ground at 3 seconds, or D.
