Let's make a diagram to visualize the problem.
In this situation, the initial vertical speed is null because the given speed is a horizontal component. To find the time spent in the air, we can use the following formula.
[tex]y=v_{0y}t+\frac{1}{2}gt^2[/tex]
Where the initial speed is null, the acceleration is gravity, and y is the height of the table.
[tex]\begin{gathered} 1.2m=0\cdot t+\frac{1}{2}\cdot(9.81\frac{m}{s^2})\cdot t^2 \\ 1.2m=4.905(\frac{m}{s^2})\cdot t^2 \\ t^2=\frac{1.2m}{4.905(\frac{m}{w^2})} \\ t\approx0.49\sec \end{gathered}[/tex]
(a) Therefore, the ball spent 0.49 seconds in the air before hitting the ground.
To find the maximum horizontal range x, we have to use the following formula.
[tex]x=v\cdot t[/tex]
Where v = 2.5 m/s and t = 0.49 seconds. Let's use these values to find x.
[tex]\begin{gathered} x=2.5(\frac{m}{s})\cdot0.49\sec \\ x\approx1.23m \end{gathered}[/tex]
(b) Therefore, the maximum horizontal range is 1.23 meters.
At last, we have to use a triple-speed to find the answer to c.
[tex]\begin{gathered} x=3\cdot2.5(\frac{m}{s})\cdot0.49\sec \\ x\approx3.68m \end{gathered}[/tex]
(c) Therefore, the ball will land 3.68 meters away from the table.
