Posted by **Anonymous** on Saturday, December 12, 2009 at 2:14pm.

A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

- physics -
**drwls**, Saturday, December 12, 2009 at 2:17pm
Get the time to fall from the height, H = 0.950 m. Call that time T. You will need to use the acceleration of gravity (g) to compute T.

H = (g/2) T^2

Haven't you seen that equation before for falling bodies?

Then use the relationship

V T = 0.352

to get V

- physics -
**Priscilla**, Saturday, December 12, 2009 at 3:37pm
A steel ball rolls with constant velocity across a tabletop 0.950m high. IT rolls off and hits the ground +0.352m horizontally from the edge of the table. How fast was the ball rolling?

GIVEN:

Vi = 0 m/s

d = 0.950 m ---> 0.352 m

a = 9.81 m/s^2

t = ?

d = Vit + 1/2 at^2

0.352 = 1/2 (9.82 m/s^2) (t)^2

0.352 = 4.905t^2

--> Divide by 4.095

square root - 0.071763507 = t^2

0.26 s = t

- physics -
**os**, Tuesday, October 20, 2015 at 8:20am
1.35 m/s

- physics -
**Anonymous**, Wednesday, November 18, 2015 at 8:32pm
8.0

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