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Newton/gravity question

  1. Jan 31, 2004 #1
    Hello and greetings to all on this forum!

    I have just begun studying physics, and use a great book called 'how things work: physics in everyday life'. I can really recommend that book to anyone!

    However, in the first chapter, i've been able to make all review questions, except for one. It's answer is probably obvious but i cannot solve the problem. Tried it for a long time, but it only gave me a headache :)

    Here it is: 'A basketball player can leap upward 0.5 m. What is his initial velocity at the start of the leap?'

    I think i have to use the fact that the earth's gravity is 9,8 m/sec2, but i don't know how.

    Any help would be greatly appreciated! Thanks in advance and have a good day everyone!!

    Gero
     
  2. jcsd
  3. Jan 31, 2004 #2

    FZ+

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    V^2 = U^2 + 2*A*S
     
  4. Jan 31, 2004 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Actually, you would need to know v= u+ at where u is the original )(upward) speed, a= -9.8 m/s2 (negative because it is downward) and t is the time. The jumper will continue upward until v= 0 so at the jumper's highest point, u-9.8t= 0. You will also need to know that d= ut+ (a/2)t2 where d is the height and, again, u is the original speed, a= -9.8 and t is the time. At the jumper's highest point, d= 0.5 so 0.5= ut- 4.9t2. From the first equation, u= 9.8t so ut= 9.8t2 and the second equation becomes 0.5= 9.8t2- 4.9t2 or
    4.9t2= 0.5. Solve that for t and then use u= 9.8t to find the original speed.
     
  5. Feb 1, 2004 #4
    As Halls said the Player will move the height just when its final velcoity is zero.
    so
    u have

    v2=u2+2as

    here a=-g and s=0.5 m
     
  6. Feb 2, 2004 #5
    Thanks to everyone! With your help i was able to solve the problem (t=0.32, u=3.136)!

    Again, thanks! Thanks to you physics continues to be fun!

    Gero
     
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