# Velocity, etc

1. Feb 13, 2009

### lauriecherie

1. The problem statement, all variables and given/known data

A ball rolls horizontally off the edge of a tabletop that is 1.40 m high. It strikes the floor at a point 1.56 m horizontally away from the table edge. (Neglect air resistance.)
(a) How long was the ball in the air?
_____ s

(b) What was its speed at the instant it left the table?
______ m/s

2. Relevant equations

3. The attempt at a solution

Is 3.12 s correct? And is 30.576 m/s correct? Seems awfully fast....?

2. Feb 13, 2009

### LowlyPion

How do you determine the time?

y = 1/2*g*t2

3. Feb 13, 2009

### lauriecherie

Ok the correct answer for time is .53 s

4. Feb 13, 2009

### LowlyPion

That's what I get.

5. Feb 13, 2009

### cjl

That is correct so far. Now, how would you determine the speed it was going at when it left the table?

It may help to think in individual x and y coordinates for this part.

6. Feb 13, 2009

### Gnosis

t = SQR(2s / a)

where,

a = acceleration (in m/s^2) per Earth’s gravity = 9.8 m/s^2
s = distance (in meters)
t = time (in seconds)
v = velocity (in m/s)
SQR = square root of the product within the parenthesis

Once you’ve derived the correct time (t), divide the 1.56 meter horizontal distance achieved by the ball, by the time (t) to derive the correct velocity (v):

1.56 meters / t = velocity (in m/s)