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

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Homework Help Overview

The problem involves a ball rolling off a table, with a focus on determining the time it is in the air and its initial velocity as it leaves the table. The context is kinematics, specifically projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the equations of motion relevant to the problem, noting the presence of two unknowns: time and initial velocity. Some question the necessity of initial vertical velocity and explore the implications of dropping the ball from a height.

Discussion Status

The discussion is ongoing, with participants exploring various equations and questioning assumptions about initial conditions. Some guidance has been offered regarding the equations to use, but no consensus has been reached on the approach to solve for the unknowns.

Contextual Notes

Participants are considering the height from which the ball falls and the absence of initial vertical velocity, which may influence their calculations. There is also a mention of homework constraints that may limit the methods available for solving the problem.

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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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