Sure! Let's solve the problem step-by-step.
We are given the quadratic model for the height of a ball:
[tex]\[ f(x) = -5x^2 + 200 \][/tex]
This equation gives the height [tex]\( f(x) \)[/tex] of the ball in meters after [tex]\( x \)[/tex] seconds.
We need to find the time [tex]\( x \)[/tex] when the height of the ball is 50 meters from the ground.
1. First, set up the equation where [tex]\( f(x) = 50 \)[/tex]:
[tex]\[ -5x^2 + 200 = 50 \][/tex]
2. Next, subtract 50 from both sides to set the equation to zero:
[tex]\[ -5x^2 + 200 - 50 = 0 \][/tex]
[tex]\[ -5x^2 + 150 = 0 \][/tex]
3. To simplify, isolate the [tex]\( x^2 \)[/tex] term by moving 150 to the other side:
[tex]\[ -5x^2 = -150 \][/tex]
4. Divide both sides by -5:
[tex]\[ x^2 = 30 \][/tex]
5. Solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\[ x = \pm\sqrt{30} \][/tex]
The solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \sqrt{30} \][/tex]
[tex]\[ x = -\sqrt{30} \][/tex]
Since [tex]\( x \)[/tex] represents time and must be non-negative, we discard the negative solution:
[tex]\[ x = \sqrt{30} \][/tex]
6. Calculate the numerical value:
[tex]\[ \sqrt{30} \approx 5.48 \][/tex]
Thus, the ball is 50 meters from the ground after about 5.48 seconds.
Therefore, the correct answer is:
[tex]\[ 5.48 \][/tex]
