A ball rolls horizontally off the edge of a tabletop that is 1.0m high. It strikes the floor at a point 1.5m horizontally away from the edge of the table. Find the speed when the ball left the table.

Find the time it takes to fall 1 meter. Then, you have that time for the horizontal velocity to act, you are given the distance, solve for the horizontal velocity

It's a two step problem dividing the problem by the coordinates x and y. The horizontal velocity (initial velocity) is 3.33m/s. 5 years late haha.

To find the speed at which the ball left the table, we can break down the problem into its horizontal and vertical components.

First, let's focus on the vertical motion. The ball falls from a height of 1.0m, and we need to calculate the time it takes to fall this distance. We can use the equation for the vertical motion:

h = (1/2)gt^2

Where:
- h is the height (1.0m in this case)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time

Rearranging the equation, we have:

t^2 = (2h)/g

Now plug in the values:

t^2 = (2 * 1.0) / 9.8
t^2 = 0.2041
t ≈ 0.45 seconds

Therefore, it takes approximately 0.45 seconds for the ball to fall 1.0 meter vertically.

Next, we can focus on the horizontal motion. The ball travels a horizontal distance of 1.5m in the same amount of time it takes to fall 1.0 meter vertically, which is approximately 0.45 seconds.

The horizontal distance traveled can be calculated using the equation:

d = vt

Where:
- d is the horizontal distance (1.5m in this case)
- v is the horizontal velocity
- t is the time (0.45 seconds)

Now we can solve for the horizontal velocity:

v = d / t
v = 1.5 / 0.45
v ≈ 3.33 m/s

Therefore, the speed at which the ball left the table is approximately 3.33 m/s.