+0  
 
0
410
8
avatar+1832 

ok thats nice I know 1 and 3 

 

just i want some one explain problem 2 for me with illustrate picture 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

physics
xvxvxv  Oct 19, 2014

Best Answer 

 #3
avatar+394 
+13

 

Part A)  From the observer’s point of view it would seem like a free fall from rest, because the ceiling, floor and person in the elevator have the same velocity. So, standard gravity acceleration would apply here.

A(Z) =-9.8m/t2* Z0 3m

Z(f) = 0

0= -0.5gt2 +Z0

T=sqr(2Z0/g)* 2*3.0m)/9.8ms2 = 0.782s

 0.782 seconds to reach the floor.

Part B) 9.8ms2*0.782 = 7.67 m/s velocity at impact with floor

Part C) Observer at relative rest.* V of Z0 = +2.5m/s (initial)

A(Z) = -9.8m/s2 * Zf =V(Z0) – gt = (2.5m/s) -(9.8m/s2) (0.782s)

Zf = -5.16m/s. Velocity is (5.16m/s) for observer at relative rest.

Part D)  The floor will rise ((2.5m/s)(0.782s)) =1.96m during the fall of the bolt.

The bolt appears to fall (3.00m -1.96m) =1.04m

The bolt will appear to fall 1.04m for observer at relative rest.

Relative reference frames are a little confusing. I can hardly wait until I get into the quantum s**t. I doubled checked this, but Melody, CPhill, and Alan should confirm this before you accept it.

 

 

 

 --7UP--

SevenUP  Oct 19, 2014
 #1
avatar+1832 
0

I found part a for problem 1 and its  25.2 m/s 

and part b for problem 1 and its 32.1 m/s 

xvxvxv  Oct 19, 2014
 #2
avatar+1832 
0

ok thats nice I know 1 and 3 

 

just i want some one explain problem 2 for me with illustrate picture 

xvxvxv  Oct 19, 2014
 #3
avatar+394 
+13
Best Answer

 

Part A)  From the observer’s point of view it would seem like a free fall from rest, because the ceiling, floor and person in the elevator have the same velocity. So, standard gravity acceleration would apply here.

A(Z) =-9.8m/t2* Z0 3m

Z(f) = 0

0= -0.5gt2 +Z0

T=sqr(2Z0/g)* 2*3.0m)/9.8ms2 = 0.782s

 0.782 seconds to reach the floor.

Part B) 9.8ms2*0.782 = 7.67 m/s velocity at impact with floor

Part C) Observer at relative rest.* V of Z0 = +2.5m/s (initial)

A(Z) = -9.8m/s2 * Zf =V(Z0) – gt = (2.5m/s) -(9.8m/s2) (0.782s)

Zf = -5.16m/s. Velocity is (5.16m/s) for observer at relative rest.

Part D)  The floor will rise ((2.5m/s)(0.782s)) =1.96m during the fall of the bolt.

The bolt appears to fall (3.00m -1.96m) =1.04m

The bolt will appear to fall 1.04m for observer at relative rest.

Relative reference frames are a little confusing. I can hardly wait until I get into the quantum s**t. I doubled checked this, but Melody, CPhill, and Alan should confirm this before you accept it.

 

 

 

 --7UP--

SevenUP  Oct 19, 2014
 #4
avatar+1832 
0

can you draw an illustrate picture ?

xvxvxv  Oct 20, 2014
 #5
avatar+26741 
+10

I agree with SevenUP's results (though I would probably have just subtracted 2.5 from 7.67 for part C).

.

Alan  Oct 20, 2014
 #6
avatar+26741 
+10

Are you looking for something like this?

 

relative observations

.

Alan  Oct 20, 2014
 #7
avatar+1832 
0

In pard d .. ehy the distance is not 3 m .. the tall of the elevator ? 

xvxvxv  Oct 21, 2014
 #8
avatar+26741 
+10

Look at the bottom two pictures. Imagine the bolt starts to drop just as the floor of the elevator is at the same level as the observer's feet  By the time the bolt hits the floor of the elevator, the floor will have moved upwards.  

 

Imagine the observer has a large vertical ruler 3m long.  He positions the top of the ruler at the height of the top of the elevator, when the elevator floor is at his feet.  The bottom of the ruler, level with the floor of the elevator at the instant the bolt starts to drop, would be 3m below the top.  Because the elevator is rising, by the time the bolt hits the elevator floor, the floor, and hence the bolt, will be at a level above the bottom of the ruler (which isn't in the elevator and so doesn't move).  This means the ruler will measure a distance less than 3m for the drop.

Alan  Oct 21, 2014

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