If two objects at rest and at the same height begin a free fall one second apart how long after the first object begins to fall will the two objects be 10 meters apart?
This is how I solve the problem:
Let t=0 be the time the first one is dropped. The distance that it falls after that is
y1 = (g/2)t^2 = 4.9 t^2
The distance that the second one :
y2 = (g/2)(t+1)^2 = 4.9 (t +1)^2
y1 - y2 = 10
4.9 t^2 - 4.9 (t +1)^2 = 10
t = 0.52s
However, the solution is different, its kind reverse from what I did. They set t, and t-1 for y1, and y2, and find t = 1.52s. I don't understand this set up, since the first one is drop first, it should be t, and the second one is drop 1 sec later, so it must be t + 1 right ? how come t - 1 ????
This is how I solve the problem:
Let t=0 be the time the first one is dropped. The distance that it falls after that is
y1 = (g/2)t^2 = 4.9 t^2
The distance that the second one :
y2 = (g/2)(t+1)^2 = 4.9 (t +1)^2
y1 - y2 = 10
4.9 t^2 - 4.9 (t +1)^2 = 10
t = 0.52s
However, the solution is different, its kind reverse from what I did. They set t, and t-1 for y1, and y2, and find t = 1.52s. I don't understand this set up, since the first one is drop first, it should be t, and the second one is drop 1 sec later, so it must be t + 1 right ? how come t - 1 ????
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First I will verify the answers and then answer to your question.
Taking the answer 1.52 second as correct,
Distance traveled by the first ball is 0.5 g t^2 = 4.9*1.52^2 = 11.32m.
Distance traveled by the second ball is 0.5 g t^2 = 4.9*0.52^2 = 1.32 m.
Distance of separation = 11.32 – 1.32 = 10 m.
Hence the answer is correct.
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Answer to your question.
The first ball is dropped at time t = 0s.
In one second it will move a distance of 0.5gt^2 = 4.9*1^2 = 4.9 m.
When the first ball is at a distance of 4.9 m from the top, the second ball is dropped.
Already a time of 1 second has elapsed.
Now the second ball is dropped. The second ball travels for 0.52second and moves distance of 0.5gt^2 = 4.9*(0.520^2 = 1.32 m
During this 0.52 s the first ball moves from the distance of 4.9 m to a distance of 11.32m.
I hope now you will be clear that the first ball has traveled for a time of 1.52 second and the second ball travels less time equal to 1.52- 1 = 0.52 second.
Thus you should use (t -1) in the equation and not t+1s.
For 1 second the second ball was at rest and actually it has traveled only for (t-1) s.
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Another example for this.
Suppose a car starts at 8 A.M.
After one hour a second car starts, i.e. at 9Am.
At 11 AM the first car has traveled a distance of 100 km.
The time of travel for the first car is 11AM – 8 AM = 3 hours.
For the second car the time of travel is 11 AM – 9AM = 2 hour.
Thus it is clear that if t = 3 hours, then for the second car the time of travel is
t -1 = 3 -1 = 2 hours.
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Taking the answer 1.52 second as correct,
Distance traveled by the first ball is 0.5 g t^2 = 4.9*1.52^2 = 11.32m.
Distance traveled by the second ball is 0.5 g t^2 = 4.9*0.52^2 = 1.32 m.
Distance of separation = 11.32 – 1.32 = 10 m.
Hence the answer is correct.
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Answer to your question.
The first ball is dropped at time t = 0s.
In one second it will move a distance of 0.5gt^2 = 4.9*1^2 = 4.9 m.
When the first ball is at a distance of 4.9 m from the top, the second ball is dropped.
Already a time of 1 second has elapsed.
Now the second ball is dropped. The second ball travels for 0.52second and moves distance of 0.5gt^2 = 4.9*(0.520^2 = 1.32 m
During this 0.52 s the first ball moves from the distance of 4.9 m to a distance of 11.32m.
I hope now you will be clear that the first ball has traveled for a time of 1.52 second and the second ball travels less time equal to 1.52- 1 = 0.52 second.
Thus you should use (t -1) in the equation and not t+1s.
For 1 second the second ball was at rest and actually it has traveled only for (t-1) s.
======================================…
Another example for this.
Suppose a car starts at 8 A.M.
After one hour a second car starts, i.e. at 9Am.
At 11 AM the first car has traveled a distance of 100 km.
The time of travel for the first car is 11AM – 8 AM = 3 hours.
For the second car the time of travel is 11 AM – 9AM = 2 hour.
Thus it is clear that if t = 3 hours, then for the second car the time of travel is
t -1 = 3 -1 = 2 hours.
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