A ball rolls off a tabletop 0.900 m above the floor and lands on the floor 2.60 m away from a point that is directly under the edge of the table. At what speed did the ball roll off of the table?

I would really appreciate it if someone could explain to me how the problem is worked out.

This ball fell during t=sqrt2h/g) =sqrt(2•0.9/9.8)=0.43 s

Horizontal motion (v(x)=const) =>
L=v(x) •t
v(x)=L/t=2.6/0.43=6.1 m/s

To solve this problem, we can use the principles of projectile motion. Here's a step-by-step explanation:

Step 1: Define the given information:
- The height of the tabletop above the floor (h) = 0.900 m
- The horizontal distance from the table's edge to the point where the ball lands (x) = 2.60 m

Step 2: Determine the vertical motion equation:
The vertical motion of the ball can be described using the equation for free fall:
h = (1/2)gt^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.

Since the ball rolls off the table horizontally, the time of flight is the same for both the horizontal and vertical components of motion.

Step 3: Find the time of flight:
Using the vertical motion equation, we can solve for t:
h = (1/2)gt^2
0.900 = (1/2)(9.8)t^2

Rearranging the equation:
t^2 = (2 * 0.900) / 9.8
t^2 ≈ 0.1837
t ≈ √0.1837
t ≈ 0.429 seconds

We have found the time of flight for the ball.

Step 4: Calculate the horizontal velocity:
To find the horizontal velocity (Vx), we can use the equation:
x = Vx * t, where x is the horizontal distance and t is the time of flight.

Rearranging the equation:
Vx = x / t
Vx = 2.60 / 0.429
Vx ≈ 6.06 m/s

Step 5: Determine the vertical velocity:
Since the ball rolls off the table horizontally, it does not have an initial vertical velocity.

Step 6: Calculate the resultant velocity:
To find the resultant velocity (V) of the ball, we can use the Pythagorean theorem:
V^2 = Vx^2 + Vy^2, where Vx is the horizontal velocity and Vy is the vertical velocity.

Since Vy is zero (due to no initial vertical velocity), we have:
V^2 = Vx^2 + 0
V = √Vx^2
V = Vx

So, the speed at which the ball rolled off the table is approximately 6.06 m/s.