Solution
- The solution steps are given below:
[tex]\begin{gathered} f(x)=-16x^2+64 \\ x\text{ is the time in seconds} \\ f(x)\text{ is the distance of the pumpkin from the ground} \\ \\ \text{ When the pumpkin is on the ground, its distance from the ground is zero. Thus,} \\ f(x)=0,\text{ when the pumpking is on the ground.} \\ \\ -16x^2+64=0 \\ \text{ Subtract 64 from both sides} \\ -16x^2=-64 \\ \text{ Divide both sides by -16} \\ x^2=-\frac{64}{-16} \\ \\ x^2=4 \\ \text{ Take the square root of both sides} \\ \\ x=\pm\sqrt{4} \\ x=+2\text{ or }-2 \\ \\ \text{ Since we cannot have negative time, } \\ x=2\text{ seconds} \end{gathered}[/tex]
Final Answer
It takes 2 seconds for the pumpkin to hit the ground
