Solving a Physics Problem: Falling Ball from Tabletop

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Homework Help Overview

The problem involves a ball rolling off a tabletop and falling to the ground, requiring calculations related to the time of flight and the speed at which the ball leaves the table. The context is kinematics, specifically projectile motion, with a focus on vertical and horizontal components of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of time in the air using the equation for vertical motion. There are attempts to verify the time of flight, with one participant suggesting a specific value and others confirming it. Questions arise regarding the speed of the ball as it leaves the table, with suggestions to analyze the problem using separate x and y coordinates.

Discussion Status

The discussion is ongoing, with some participants confirming the time of flight as approximately 0.53 seconds. Guidance is provided on how to approach the calculation of the horizontal speed, indicating a productive exploration of the problem.

Contextual Notes

Participants are working under the assumption that air resistance can be neglected, and they are focused on deriving values based on the given height and horizontal distance.

lauriecherie
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Homework Statement



A ball rolls horizontally off the edge of a tabletop that is 1.40 m high. It strikes the floor at a point 1.56 m horizontally away from the table edge. (Neglect air resistance.)
(a) How long was the ball in the air?
_____ s

(b) What was its speed at the instant it left the table?
______ m/s

Homework Equations





The Attempt at a Solution



Is 3.12 s correct? And is 30.576 m/s correct? Seems awfully fast...?
 
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How do you determine the time?

y = 1/2*g*t2
 
Ok the correct answer for time is .53 s
 
lauriecherie said:
Ok the correct answer for time is .53 s

That's what I get.
 
That is correct so far. Now, how would you determine the speed it was going at when it left the table?

It may help to think in individual x and y coordinates for this part.
 
t = SQR(2s / a)

where,

a = acceleration (in m/s^2) per Earth’s gravity = 9.8 m/s^2
s = distance (in meters)
t = time (in seconds)
v = velocity (in m/s)
SQR = square root of the product within the parenthesis

Once you’ve derived the correct time (t), divide the 1.56 meter horizontal distance achieved by the ball, by the time (t) to derive the correct velocity (v):

1.56 meters / t = velocity (in m/s)
 

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