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Projectile motion: equal distances in 5th and 6th seconds of trajectory

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    With what speed should a body be thrown upward so the distance traversed in the 5th second and 6th second are equal?

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

    3. The attempt at a solution
    This doesn't make sense to me, the distance travelled 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.
     
  2. jcsd
  3. Mar 20, 2017 #2

    jedishrfu

    Staff: Mentor

    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?
     
  4. Mar 20, 2017 #3
    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.
     
  5. Mar 20, 2017 #4

    gneill

    User Avatar

    Staff: Mentor

    Moderator note: Thread title was too general (which is contrary to forum rules). Changed it to reflect the actual problem.
     
  6. Mar 20, 2017 #5
    Am I throwing the ball vertically up , or at angle ?
     
  7. Mar 20, 2017 #6
    "Upward" in your problem statement means vertically up.
     
  8. Mar 20, 2017 #7
    Vf = Vi - gt
    0= V
    i - 10(1)
    V
    i = 10

    Is this correct ?
     
  9. Mar 20, 2017 #8

    cnh1995

    User Avatar
    Homework Helper

    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?
     
  10. Mar 20, 2017 #9
    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.
     
  11. Mar 20, 2017 #10
    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
    V
    f = 0
    V
    i = gt= 10(5)= 50.
    Am I right ?
     
  12. Mar 20, 2017 #11

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good.
     
  13. Mar 20, 2017 #12
    Units? :oldtongue:
     
  14. Mar 20, 2017 #13
    Light year per second.
     
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