Solve for the time and initial velocity of a ball rolling off a table?"

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In summary, the problem involves a ball rolling off a table and landing 1.78 m away from the edge of the table. The equations x=vcos(x)t and y=vsin(x)t- .5gt^2 can be used to solve for the time the ball is in the air and its speed at the moment it leaves the table. Another equation, y=ut+\frac{1}{2}gt^2, can be used to solve for the time it takes for an object to fall from a height of 1.19 m. Without any initial vertical velocity, the time it takes for the ball to fall is approximately 0.493 seconds.
  • #1
TS656577
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[SOLVED] Physics again

Homework Statement



A small ball rolls horizontally off the edge of a tabletop that is 1.19 m high. It strikes the floor at a point 1.78 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Homework Equations


x=vcos(x)t
y=vsin(x)t- .5gt^2


The Attempt at a Solution



I believe those are the right equations that I need, but I have two unknowns in the equation, velocity and time.
 
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  • #2
The only other equation that I could maybe see using is y=(tan(*)x - (gx^2)/2(vcos(*)^2)
 
  • #3
If the ball were simply dropped from a height of 1.19 m, how long would it take to fall?
 
  • #4
This may sound stupid, but I have no idea...
 
  • #5
TS656577 said:
This may sound stupid, but I have no idea...

[tex]y=ut+\frac{1}{2}gt^2[/tex]
(u=initial velocity)

Vertically...what is the height it falls from? Is there any initial vertical velocity?
 
  • #6
Hmmm since there is no initial velocity, then the "ut" in essence cancels out to 0, thus, solving for t I got, .493 seconds.
 

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