ANSWER
0.89 m
EXPLANATION
Assuming that the inclined plane is frictionless, the energy must be conserved.
If the marble starts from rest, then at the top of the inclined plane it has only gravitational potential energy. Then, at the end of the inclined plane, the marble will have only kinetic energy if we consider that the table is the reference. This energy must be the same as the gravitational potential energy the marble had at the top of the inclined plane, because of the Law of Conservation of Energy:
[tex]PE_i=KE_f[/tex]
Replace with the known equations for each kind of energy,
[tex]m\cdot g\cdot h=\frac{1}{2}\cdot m\cdot v^2[/tex]
From this equation, we can find the velocity of the marble when it leaves the table. The mass of the marble, m, cancels out,
[tex]v=\sqrt[]{2\cdot g\cdot h}[/tex]
The height, h, is the distance from the table to the inclined plane, 0.26 m. The acceleration due to gravity is 9.81 m/s²,
[tex]v=\sqrt[]{2\cdot9.81m/s^2\cdot0.26m}\approx2.2586m/s[/tex]
Now, from the diagram, we can assume that the marble leaves the table horizontally, with an initial velocity of approximately 2.2586 m/s,
Note that the path the marble will follow is a projectile motion path. The height of a projectile is,
[tex]y=y_o+v_{oy}t-\frac{1}{2}gt^2[/tex]
The marble is thrown horizontally, so there is no initial vertical velocity. We can find the time it takes for the marble to hit the ground, which is when y = 0,
[tex]0=0.76m-\frac{1}{2}\cdot9.81m/s^2\cdot t^2[/tex]
Solving for t,
[tex]t=\sqrt[]{\frac{2\cdot0.76m}{9.81m/s^2}}\approx0.3936s[/tex]
The horizontal distance a projectile travels is given by,
[tex]dx=v_{ox}t[/tex]
There is no horizontal acceleration and the horizontal velocity is constant, which is the initial velocity of the marble, in this case,
[tex]dx=2.2586m/s\cdot0.3936s\approx0.89m[/tex]
Hence, the marble will hit the floor at approximately 0.89 m from the table.
