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

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The problem involves a ball rolling off a 1.19 m high table and landing 1.78 m away. To determine the time the ball is in the air, the vertical motion equation is used, leading to a calculated fall time of approximately 0.493 seconds. The horizontal motion equation is applied to find the initial velocity, which is dependent on the time in the air. The equations used include y = ut + 0.5gt² and x = vcos(θ)t. The discussion emphasizes solving for both time and initial velocity using the principles of projectile motion.
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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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The only other equation that I could maybe see using is y=(tan(*)x - (gx^2)/2(vcos(*)^2)
 
If the ball were simply dropped from a height of 1.19 m, how long would it take to fall?
 
This may sound stupid, but I have no idea...
 
TS656577 said:
This may sound stupid, but I have no idea...

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

Vertically...what is the height it falls from? Is there any initial vertical velocity?
 
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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