Answer:
The magnitude of the acceleration of the ball is 9.81m / ^ 2 down, as you can see this value corresponds to gravity.
Explanation:
This movement is known as semi-parabolic, consider the following to solve
1) horizontally (X) there is a constant movement, with zero acceleration.
2) vertically (Y) there is a constant acceleration normally of the gravity value.
3) from the moment the ball begins to fall until it touches the ground the gravity is the same, therefore the acceleration is the same.
4) the initial vertical speed is 0.
When performing a mathematical demonstration, it is found that the equation that define this motion as the follows
[tex]Y=VoT+0.5at^{2}[/tex]
where
Y= table height=0.75m
Vo=0=initial speed
t=time=0.391s
a=acceleration
Solving
[tex]Y=0.5at^2\\a=\frac{Y}{(0.5)(t^2)} \\a=\frac{0.75m}{(0.5)((0.391)^2)} =9.81m/s^2[/tex]
The magnitude of the acceleration of the ball is 9.81m / ^ 2 down, as you can see this value corresponds to gravity.
