ANSWER
28.56 m
EXPLANATION
Let's make a diagram of this situation to understand it better,
The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.
We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,
[tex]\Delta x=v_{ox}\cdot t[/tex]
The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,
[tex]\Delta y=v_{oy}\cdot t+\frac{1}{2}gt^2[/tex]
The initial vertical velocity is zero because the ball is hit horizontally,
[tex]\Delta y=\frac{1}{2}gt^2[/tex]
Solve for t. Multiply both sides by 2/g,
[tex]\begin{gathered} \Delta y\frac{2}{g}=\frac{1}{2}g\cdot\frac{2}{g}\cdot t^2 \\ \frac{2\Delta y}{g}=t^2 \end{gathered}[/tex]
And take the square root to both sides of the equation,
[tex]t=\sqrt[]{\frac{2\Delta y}{g}}[/tex]
Δy is the height of the hill, 2.5m, and g = 9.81m/s²,
[tex]t=\sqrt[]{\frac{2\cdot2.5m}{9.81m/s^2}}=\sqrt[]{\frac{5m}{9.81m/s^2}}=\sqrt[]{0.509684s^2}=0.713922s[/tex]
This is the time the ball was in the air for. Now we can find the distance it traveled,
[tex]\Delta x=40m/s\cdot0.713922s=28.56m[/tex]
The golf ball flew 28.56 m horizontally.
