Solving a Physics Problem: Falling Ball from Tabletop

AI Thread Summary
The problem involves a ball rolling off a 1.40 m high tabletop and landing 1.56 m away horizontally. The time the ball was in the air is calculated to be 0.53 seconds, using the equation y = 1/2 * g * t². To find the speed at which the ball left the table, the horizontal distance is divided by the time in the air. This results in a velocity of approximately 2.94 m/s. Understanding the separation of x and y coordinates is crucial for solving such projectile motion problems.
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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