To determine the initial velocity, \( v_0 \), of the golf ball, we can use the equation that models its height:
[tex]\[ y = -16x^2 + v_0 x \][/tex]
Given that the ball returns to the ground 3 seconds after it is hit, the height \( y \) is 0 when \( x = 3 \) seconds. Let's substitute these values into the equation to solve for the initial velocity \( v_0 \):
[tex]\[ y = 0 \][/tex]
Substitute \( x = 3 \):
[tex]\[ 0 = -16(3)^2 + v_0 (3) \][/tex]
Simplify the equation step by step:
[tex]\[ 0 = -16 \times 9 + 3 v_0 \][/tex]
[tex]\[ 0 = -144 + 3 v_0 \][/tex]
Now, isolate \( v_0 \) by adding 144 to both sides of the equation:
[tex]\[ 144 = 3 v_0 \][/tex]
Finally, solve for \( v_0 \) by dividing both sides by 3:
[tex]\[ v_0 = \frac{144}{3} \][/tex]
[tex]\[ v_0 = 48.0 \][/tex]
Therefore, the initial velocity of the golf ball is:
[tex]\[ v_0 = 48.0 \text{ feet per second} \][/tex]