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The function [tex]$y=-16 x^2+v_0 x$[/tex] models the height of a golf ball in feet [tex]x[/tex] seconds after a player hits it. In the equation of the function, [tex]v_0[/tex] is the initial velocity of the ball in feet per second. The ball returns to the ground 3 seconds after it is hit.

What is the value of [tex]v_0[/tex]?

Sagot :

GinnyAnswer
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]
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