Why is distance the same in the 5th and 6th seconds of upward motion?

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NoahCygnus
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Homework Statement


With what speed should a body be thrown upward so the distance traversed in the 5th second and 6th second are equal?

Homework Equations


No idea since I fail to comprehend the statement of the problem.

The Attempt at a Solution


This doesn't make sense to me, the distance traveled during the fifth second should be more than the sixth second as the initial velocity in the fifth second is more. During the sixth second , final velocity of fifth second will be the initial velocity of sixth second and it should be less than the initial velocity of the fifth second , so displacement should be different.
 
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You throw a ball in the air, it goes up, stops momentarily and comes back down. At the moment it stopped its speed is zero, one second prior it was X and one second after it will be X too. What is X?

Speed has no direction its a scalar, velocity has direction but they didn't ask that in your question.

As a clue ask yourself what time is the speed zero?
 
Note that when they say "the distance traversed", they mean the distance to the body from a fixed point (say, from where the body was launched). You're perhaps confusing it with the total length of the curve the body traces out (trajectory) as it's airborne. You should be able to solve it with jedishrfu's hint.
 
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Moderator note: Thread title was too general (which is contrary to forum rules). Changed it to reflect the actual problem.
 
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jedishrfu said:
You throw a ball in the air, it goes up, stops momentarily and comes back down. At the moment it stopped its speed is zero, one second prior it was X and one second after it will be X too. What is X?

Speed has no direction its a scalar, velocity has direction but they didn't ask that in your question.

As a clue ask yourself what time is the speed zero?
jedishrfu said:
You throw a ball in the air, it goes up, stops momentarily and comes back down. At the moment it stopped its speed is zero, one second prior it was X and one second after it will be X too. What is X?

Speed has no direction its a scalar, velocity has direction but they didn't ask that in your question.

As a clue ask yourself what time is the speed zero?
Am I throwing the ball vertically up , or at angle ?
 
NoahCygnus said:
Am I throwing the ball vertically up , or at angle ?
"Upward" in your problem statement means vertically up.
 
jedishrfu said:
You throw a ball in the air, it goes up, stops momentarily and comes back down. At the moment it stopped its speed is zero, one second prior it was X and one second after it will be X too. What is X?

Speed has no direction its a scalar, velocity has direction but they didn't ask that in your question.

As a clue ask yourself what time is the speed zero?
jedishrfu said:
You throw a ball in the air, it goes up, stops momentarily and comes back down. At the moment it stopped its speed is zero, one second prior it was X and one second after it will be X too. What is X?

Speed has no direction its a scalar, velocity has direction but they didn't ask that in your question.

As a clue ask yourself what time is the speed zero?
Vf = Vi - gt
0= Vi - 10(1)
Vi = 10

Is this correct ?
 
NoahCygnus said:
Vf = Vi - gt
0= Vi - 10(1)
Vi = 10

Is this correct ?
Distance covered from t=4s to t=5s is equal to the distance covered from t=5s to t=6s.

What can you say about the direction of motion in both the intervals?
 
cnh1995 said:
Distance covered from t=4s to t=5s is equal to the distance covered from t=5s to t=6s.

What can you say about the direction of motion in both the intervals?
After fifth second , the motion gets reversed and the object falls ?
So do I've to calculate initial velocity at the fourth second ? I still don't get it.
 
NoahCygnus said:
After fifth second , the motion gets reversed and the object falls ?
So do I've to calculate initial velocity at the fourth second ? I still don't get it.
I think I get it now. The final velocity of the body should become 0 at fifth second. As the body will fall after fifth second and projectile motion is symmetric, the distance covered during 4 to 5th second and 5th to 6th second should be the same. So the initial velocity/speed with which I throw the body upward should be such that the body reaches max height at fifth second .
Vf = Vi - gt
Vf = 0
Vi = gt= 10(5)= 50.
Am I right ?
 
NoahCygnus said:
I think I get it now. The final velocity of the body should become 0 at fifth second. As the body will fall after fifth second and projectile motion is symmetric, the distance covered during 4 to 5th second and 5th to 6th second should be the same. So the initial velocity/speed with which I throw the body upward should be such that the body reaches max height at fifth second .
Vf = Vi - gt
Vf = 0
Vi = gt= 10(5)= 50.
Am I right ?

Looks good.
 
NoahCygnus said:
I think I get it now. The final velocity of the body should become 0 at fifth second. As the body will fall after fifth second and projectile motion is symmetric, the distance covered during 4 to 5th second and 5th to 6th second should be the same. So the initial velocity/speed with which I throw the body upward should be such that the body reaches max height at fifth second .
Vf = Vi - gt
Vf = 0
Vi = gt= 10(5)= 50.
Am I right ?
Units? :oldtongue:
 
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