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Homework Help: Anyone awake to answer questions?

  1. Dec 14, 2004 #1
    I need to know if anyone is up to answer more questions... They aren't very hard. Just questions from the first semester of an A.P Physics Course. If you are able to answer questions...pls respond to this and I will create a new thread for the questions and anyone else who needs the review can join! All help is greatly appreciated... I'm a smart student I've just had a very rough year and im sitting with a 64 average in this class and I have a final multiple choice exam and a make-up test tomorrow and if I score high, I can pass. I love physics, mainly Quantum Physics, so far I only know the concepts and such of Quantum Physics to a limited basis so I thought a college level physics course would help me be able to further my knowledge faster. So anyone who can help me, could be helping a person that could change the world in Quantum Mechanics and Quantum Computing in the future as I pursue my dream!!
  2. jcsd
  3. Dec 14, 2004 #2
    Instead of wasting time asking to ask, just ask it, and hope someone would answer :)
  4. Dec 14, 2004 #3
    well i didnt wana ask a question and have to wait till later to get the answer because id have to have it in the next 5 hours when I leave for school
    ok the next post I make will have my question.
  5. Dec 15, 2004 #4
    A football is kicked at an angle (theta) with respect to the horizontal. Which one of the following statements best describes the acceleration of the football during this event if air resistance is neglected?

    A) The acceleration is 0 m/s^2 at all times.
    B)The acceleration is 9.8 m/s^2 at all times
    C)The acceleration is zero when the football has reached the highest point in its trajectory.
    D)The acceleration is positive as the football rises, and its negative as the football falls.
    E)The acceleration starts at 9.8 m/s^2 and drops to some constant lower value as the ball approaches the ground.

    The answer the teacher gave me is B, but to me it seems like D. I could have messed up when copying. Could you tell me which is right and if the teacher is right why his is right and all the other choices are wrong?
  6. Dec 15, 2004 #5
    is it because of the force of gravity is always acting upon it with 9.8 m/s^2 and when the ball is kicked only the velocity is affected?
  7. Dec 15, 2004 #6
    B is right. Remember, without air resistance, while the football is in the air, the only force acting on it is gravity. As a result, the football is in freefall. Therefore, the acceleration is constant, and its value is 9.8m/s^2 downwards.
  8. Dec 15, 2004 #7
    So anytime an object is moving through the air without air resistance the only force of acceleration is gravity? and if air resistance is their, that is the only thing that will affect the acceleration?
  9. Dec 15, 2004 #8
    If there is no external or resistive force, yes.

    For instance, an airplane or a rocket.. has a force acting on it while it's flying.. so obviously it has a net force/net acceleration.

    But if you're only throwing a ball through the air... then what you said is right :-)
  10. Dec 15, 2004 #9


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    Homework Helper

    Your teacher was right and you are right. Gravity always acts on the same way. The acceleration of a body in free fall near the surface of the earth is always "g", downward.
    It is the vertical component of velocity that becomes zero at the highest point of trajectory. And it is this vertical velocity-component again, that is positive at the beginning and negative when the ball falls downward.

    But: During that short time interval while the ball is kicked, there is an other force also acting on it. The resultant acceleration during this very short time is not g. So the ball gains some velocity, and after leaning the kicking foot, it moves by "itself" with downward acceleration equal to g.

    On the same way, there are two forces acting upon the ball after it reached the ground: gravity and the normal force from the horizontal surface it touches. The two forces cancel, the acceleration becomes zero.

  11. Dec 15, 2004 #10
    alright. good now i understand something else.(My teacher is stupid and should keep teaching just normal highschool physics and not try this stuff all he does is draw pictures of things from the problems and spends so much time makeing them look good we don't learn as much as we should.)

    alright, next problem.

    A tennis ball is thrown from ground level with velocity Vo directed 30 degrees above the horizontal. If it takes the ball 1.0s to reach the top of its trajectory, what is the magnitude of the initial velocity?

    The answer is 19.6 m/s^2. I dont see how its true.

    The teacher says list what I know first and dont know.

    Thrown 30 degrees above horizontal.
    Time = 1 second.
    Initial Velocity = ?

    With this limited info, I dont see how he gets this. From my knowledge, I would think it is 0 because it starts at ground level and it has to be thrown so it would start at rest.
  12. Dec 15, 2004 #11
    First off, initial velocity = velocity at which it leaves the hand. That is just how it is defined. With that in mind...

    You know that it takes one second to reach the top of its trajectory. As a result, you also know that it takes one second to fall back to the ground. So you know that the vertical velocity is just V_y = gt = 9.8m/s.

    But the vertical velocity can also be expressed by Vo*Sin(30 deg) = Vo/2.

    From this, you can easily find that Vo = 19.6 m/s (and not m/s^2, as you indicated)
  13. Dec 15, 2004 #12
    ah, so since it takes one second to go up, another to go down... and its decelerating at 9.8 meters a second... it has to have enough acceleration to lose 9.8 meters going up, and 9.8 going down,

    what does V_y mean?
  14. Dec 15, 2004 #13
    Velocity, y component. (Vertical component of velocity)
  15. Dec 15, 2004 #14
    A quarter back throws a pass at an angle of 35degrees above the horizontal with an initial speed of 25 m/s. The ball is caught by the reciever 2.55 seconds later. Determine the distance the ball was thrown.

    My answer: (25m/s * Cos(30)) * 2.55 s
    Is it correct? My calculator just died and the correct answer is 52 m.

    An arrow is shot horizontally from a height of 4.9 m above the ground. The initial speed of the arrow is 45 m/s. Neglecting friction, how long will it take the arrow to hit the ground.

    My answer: y=Vi*t+(1/2)at^2 = time
    And is this correct? It should be 6.0s
  16. Dec 15, 2004 #15
    Your first answer is correct (I assume there is a typo n the argument of your cosine 30->35 degrees). About your second answer, if it is an answer. If you fill in the right numbers it can yield the correct answer, But you have to remember you only have to look at the vertical motion, then what is Vi?

    Or to elaborate some more. ignoring friction the horizontal movement of an object in free fall does not effect its vertical motion, it does not matter at what speed in the horizontal direction the arow moves for the time it takes to reach the ground. This is because the acceleration a=g=9,8m/s^2 is the same downwards...
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