Catching a glass falling to the ground

[AirForceOne]

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.

Homework Equations

The Attempt at a Solution

I find the position vector of the hand to be:

I get the position vector of the glass to be:

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

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?!

Please help asap!

 

[PhanthomJay]

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

 

[NewtonianAlch]

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.

 

[AirForceOne]

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.

 

[PhanthomJay]

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.