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Some quite confusing problems

  1. Jun 14, 2003 #1
    Say there's a ball being thrown upwards (from the bottom to the top of a building) with an initial speed of 29.4 m/s for 4 seconds at an acceleration of -9.8 m/s².

    Now here's a few of my problems:

    1.) My book says the ball would stop after 3 seconds. Wouldn't it therefore be pointless to calculate what would happen in 4 seconds?

    2.) My book says the end speed (V1/Vf) would be -9.8 m/s. Doesn't the end speed always equal acceleration times time (V1 = at) which would therefore be -39.2 m/s?

    Also, what's the difference between starting at rest (with the initial speed being zero) and being thrown (for example, in this case, 29.4 m/s), isn't gravity technically 'throwing' the object in question?

    Thanks to anyone who can explain this, it's confusing me alot. I've only been working with physics (self-teaching) for about 4 days so excuse the newbieness of the question.
     
  2. jcsd
  3. Jun 14, 2003 #2
    I don't think the question is perfectly clear. What you've got though, is a ball thrown upwards at 29.4 m/s, and this is all you need to find that it reaches the top of it's flight, or stops at t = 3 seconds. So the question wants you to find what it's velocity is after 4 seconds - in other words when it has been coming down again for 1 second, which is fairly trivial.

    I don't really understand what you're asking with your last question. Here, you are throwing the ball upwards so you have to give it an initial velocity or it wont go anywhere - gravity acts downwards on the ball to slow it down and eventually stop it. Can you clarify a bit if you want a further/better explanation?
     
  4. Jun 14, 2003 #3
    I agree with mulder that ur question is not really clear, i am sure that if u make it a little clearer we would be able to help you.
    But at the moment, i hope this equation till help you :
    Vf=Vi+at
     
  5. Jun 15, 2003 #4
    Thanks STAii, that pretty much cleared it up. I was using the wrong formula to calculate 'Vf'. As for my second question, I think I understand it now.

    Thanks guys.
     
  6. Jun 15, 2003 #5

    HallsofIvy

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    "1.) My book says the ball would stop after 3 seconds. Wouldn't it therefore be pointless to calculate what would happen in 4 seconds?"

    The ball starts moving upward. Since the acceleration due to gravity is downward, the ball slows down and after 3 seconds comes to a stop. (the acceleration is -9.8 m/s^2. After 3 seconds, the total "change in speed" will be (3s)(-9.8m/s^2)= -29.4 m/s so that the original speed of +29.4 m/s has reduced to 29.4- 29.4= 0 m/s.

    But the acceleration still there! The ball doesn't "hang in the air". It still has an acceleration (because gravity is still pulling on it) of -29.4 m/s^2 so after 1 MORE second, it is moving downward at -9.8 m/s. You could have calculated that without worrying about when the ball "stopped" by saying, "since the acceleration is -9.8 m/s, in 4 seconds the change in speed will be (4)(-9.8)= 39.2 m/s. Since the initial speed was 29.4 m/s, after 4 seconds, the speed will be 29.4- 39.2= -9.8 m/s."

    "2.) My book says the end speed (V1/Vf) would be -9.8 m/s. Doesn't the end speed always equal acceleration times time (V1 = at) which would therefore be -39.2 m/s?"

    Well, first, the end speed is "Vf", not "V1/Vf". I presume you mean V1 to be the initial speed. As long as acceleration is a constant, the CHANGE in speed is "acceleration times time": 4(-9.8)=
    -39.2 m/s. However, the final speed will be "initial speed plus change in speed":in this case, 29.4- 39.2= -9.8m/s as we got above.

    IF we just dropped the ball from the top of the building, the initial speed would be 0 ("dropped" not "threw up/down") and the speed after 4 seconds would be 0- 39.2= -39.2 m/s.

    "Also, what's the difference between starting at rest (with the initial speed being zero) and being thrown (for example, in this case, 29.4 m/s), isn't gravity technically 'throwing' the object in question?"

    No, gravity doesn't "throw" anything, it only "pulls"! The "acceleration due to gravity" gives you the change in speed. That has to be added to the initial speed to get the final speed.

    If you stand on the roof of the building a DROP the ball (initial speed 0), after 4 seconds its speed will be 0- 39.2= -39.2m/s and the ball will be well below the roof (if it hasn't already hit the ground!).
    If you throw the ball upward from the roof at an intitial speed of
    29.4 m/s, the CHANGE in speed after 4 seconds will be exactly the same: -39.2 m/s so its speed have decreased from 29.4: 29.4- 39.2 m/s
    = -9.8 m/s and the ball will still be well above the roof.
     
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