To determine when the golf ball hits the ground, we can use the equation of motion for a freely falling object. The formula is:
\[ h = v_0t - \frac{1}{2}gt^2 \]
Where:
- \( h \) is the height (in this case, 24 ft).
- \( v_0 \) is the initial velocity (assuming John hits the ball vertically, this is 0).
- \( g \) is the acceleration due to gravity (approximately \( 32 \, \text{ft/s}^2 \)).
- \( t \) is the time.
In this case, since John is hitting the ball vertically, the initial velocity \( v_0 \) is 0, and we're solving for \( t \) when the ball hits the ground (\( h = 0 \)).
\[ 0 = -\frac{1}{2}gt^2 + 24 \]
Now, solve for \( t \):
\[ \frac{1}{2}gt^2 = 24 \]
\[ gt^2 = 48 \]
\[ t^2 = \frac{48}{g} \]
\[ t = \sqrt{\frac{48}{g}} \]
Now plug in the value of \( g \) (approximately \( 32 \, \text{ft/s}^2 \)):
\[ t = \sqrt{\frac{48}{32}} \]
\[ t = sqr{3}{2}}
So, \( t \) is the square root of 3/2 seconds. The exact numerical value can be calculated, but this represents the time it takes for the golf ball to hit the ground from the second level of the driving range.