Catching a glass falling to the ground

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Homework Help Overview

The problem involves a scenario where a glass falls from a table after a person trips, and the individual attempts to catch it with a reaction time constraint. The discussion centers around calculating the distance the glass falls during the reaction time and determining the necessary acceleration of the hand to catch the glass before it hits the ground.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the time it takes for the glass to fall and how to calculate the distance fallen using gravitational equations. There are attempts to clarify the equations used and the assumptions regarding units of measurement. Some participants question the calculations related to the acceleration needed for the hand to reach the glass.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on the calculations and suggesting simpler approaches. There is no explicit consensus, but some guidance has been offered regarding the use of gravitational equations and the importance of unit consistency.

Contextual Notes

Participants note the difference in gravitational acceleration units between metric and imperial systems, which may affect the calculations. There is also mention of the reaction time and its impact on the time available to catch the glass.

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Homework Statement



A person trips against a table, causing a glass to fall off the edge. An excellent human reaction time is 0.25 seconds. In that time, how far wil the glass fall? Assume that immediately (after 0.25 s has elapsed) begin to acclerate your hand so that you grab the glass when it is 6 in from the floor. What constant acceleration was necessary, and how fast was your hand traveling when you contacted the glass? Assume your hand moved in a straight line.
96DN6.png

Homework Equations


The Attempt at a Solution



I find the position vector of the hand to be:
81ZwS.jpg

I get the position vector of the glass to be:
VZQmD.jpg

I need to find the time it takes for the glass to fall to y=0.5:
Ybggo.jpg

So the hand only has 0.46 seconds to reach the glass. However, when I plug this into the position vector of the hand to find a_x and a_y and find the magnitude of the acceleration, I get like twenty-something ft/s^2 which is far below the correct answer of 241 ft/s^2. What am I doing wrong?!
UtBcm.jpg


Please help asap!
 
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Good afternoon, Mr. President.

Your equations, sir, with all due respect, made me think I was looking at Wiki.

Keep it simple. Find the time it takes for the glass to be 6 inches from the floor, using distance = 1/2gt^2. Then that time less the reaction time is the time that your hand has to reach the glass, traveling a diagonal distance as found from Pythagorus' theorem.

Warmest regards,
Jay
 
I agree, keep it simple.

1) How far will the glass fall in 0.25 seconds?

You know the value of gravity (9.81 m/s^2), and you know the time (0.25 seconds). Use the equation:

S_{y} = ut + \frac{at^2}{2}

The second part has already been answered.
 
PhanthomJay said:
Good afternoon, Mr. President.

Your equations, sir, with all due respect, made me think I was looking at Wiki.

Keep it simple. Find the time it takes for the glass to be 6 inches from the floor, using distance = 1/2gt^2. Then that time less the reaction time is the time that your hand has to reach the glass, traveling a diagonal distance as found from Pythagorus' theorem.

Warmest regards,
Jay

But that's what I did...
y=(1/2)gt^2
t=sqrt(2y/g)
=sqrt((2)(2.5)/9.8)
= 0.71

0.71-0.25 = .46...

Diagonal distance=2.5,
y=(1/2)at^2
a=(2y)/(t^2)
=(2*2.5)/(0.46^2)
= 23.63ft/s^2...which is what I got before.
 
Oh, sure, that's a lot better than all those i's and j's.. But in the USA system, g is not established in units of m/sec^2, but rather, in units of ft/sec^2.


9.8m/sec^2 = appx. 32.2 ft.sec^2.:frown:
 

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