How fast was it rolling on the table before it fell off?

Question

A 78g autographed baseball rolls off of a 1.1 m high table and strikes the floor a horizontal distance of 0.65 m away from the table.


The acceleration of gravity is 9.81 m/s(squared)


how fast was it rolling on the table before it fell off? Answer in unites of m/s

Answer 1

Before the ball fell off the table, the ball had only motion in the x-direction only. When it reached the end of the table, then, it had a velocity of 0 m/s in the y-direction.

At that point, it started in 2-dimensional motion,
in the y-direction: y = y0 + vy0 * t - (1/2) * 9.81 * t^2
in the x-direction: x = x0 + vx0 * t

We are solving for "vx0", and we know that

.65 = 0 + vx0 * t

Find "t" and you find "vx0"


We also know that

0 = 1.1 + 0 * t - (1/2) * 9.81 * t^2,
since the ball had vy0 = 0 and the table is 1.1 m above the ground, and the ball reaches the ground at y = 0 m, and when x = .65 m

1.1 = (1/2) * 9.81 * t^2,
t = 0.47 sec

and

0.65 = 0 + vx0 * (0.47)
vx0 = 0.65 / 0.47

vx0 = 1.37 m/s

Answer 2

1) do your own homework.

2) you do it like this:

use the equation of motion which gives you acceleration, distance, and initial velocity. the last variable will be time. you've got gravity's acceleration, distance (height of table) and initial velocity (0, because you're talking in the Y direction). solve for time then. Now that you have time, solve by taking the distance from the table (0.65 m) and divide by that time. Tada, thats your x oriented velocity.