To solve this problem, we can use the equations of motion for an object in free fall with constant acceleration due to gravity (g = 9.81 m/s^2), which are:
y = vit + (1/2)gt^2 (Equation 1)
v = vi + gt (Equation 2)
where y is the vertical displacement, vi is the initial vertical velocity (which is zero when the marble leaves the table), t is the time, and v is the final velocity.
A. To find the time the marble is in the air, we can use Equation 1 with y = 1.1 m and vi = 0:
1.1 = (1/2)gt^2
Solving for t, we get:
t = sqrt(2*1.1/g) = 0.47 seconds
Therefore, the marble is in the air for 0.47 seconds.
B. To find the speed of the marble when it leaves the table, we can use Equation 2 with vi = 0 and t = 0.47 seconds:
v = gt = 9.810.47 = 4.61 m/s
Therefore, the speed of the marble when it leaves the table is 4.61 m/s.
C. To find the speed of the marble when it hits the floor, we need to find the horizontal and vertical components of its velocity. The horizontal component is constant, since there is no horizontal acceleration. The vertical component can be found using Equation 2 with vi = 0 and t = 0.47 seconds:
vy = gt = 9.810.47 = 4.61 m/s
The horizontal component of the velocity can be found from the distance traveled in the horizontal direction (2.1 m) and the time in the air (0.47 seconds):
vx = 2.1/0.47 = 4.47 m/s
The speed of the marble when it hits the floor is the magnitude of its velocity, which can be found using the Pythagorean theorem:
v = sqrt(vx^2 + vy^2) = sqrt(4.47^2 + 4.61^2) = 6.35 m/s
Therefore, the speed of the marble when it hits the floor is 6.35 m/s.