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September 2, 2014

September 2, 2014

Posted by **Claudia** on Friday, October 3, 2008 at 8:32pm.

a 71g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of .83m away from the table.

1.how fast was it rolling on the table before it fell off?

ok well i already figured out the first question. which was 1.61

but for some reason i keep getting the

2nd question wrong.

can someone help!please

2.What was the ball's velocity just before it hit the floor?

That is, at what angle in the range -90° to +90° relative to the horizontal directed away from the table) did the ball hit the floor?

answer in units of °

- PHYSICS -
**Damon**, Friday, October 3, 2008 at 8:51pmIt is still going 1.61 m/s horizontal if it was going that speed originally.

Vertical the speed is:

v = Vo + a t, the initial speed down, Vo, is zero

v = 0 - 9.8 t

what is t, the time to fall 1.3 m?

h = ho + Vo t + .5 a t^2

-1.3 = 0 + 0 - 4.9 t^2

t = .515 s

so

v = -9.8 * .515 = -5.05 m/s

speed = sqrt (-5.05^2 + 1.61^2)

= sqrt ( 25.5+2.59)

=5.3 m/s

tan angle to horizontal = 5.05/1.61 = 3.14

angle down from horizontal = 72.3 degrees down from horizontal

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