Someone give me a hint been at it for hrs

  • Thread starter sjcc
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In summary: Graphs of motion are used to illustrate the motion of an object. They can be used to show the change in velocity over time, the change in acceleration over time, and the change in position over time. They can be used to determine the final velocity and momentum of an object.
  • #1
sjcc
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Homework Statement



A ball is kicked from a location < 7, 0, -6 > (on the ground) with initial velocity < -9, 15, -3 > m/s. The ball's speed is low enough that air resistance is negligible.
What is the velocity of the ball 0.4 seconds after being kicked? (Use the Momentum Principle!)


Homework Equations





The Attempt at a Solution


 
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  • #2
What are your thoughts on the problem? :)
 
  • #3
i've tried dividing the location of the ball by the 0.4 seconds. it was wrong. I thought of next to multiply the location by 0.4. but I am not sure if that is correct. i think there is something that needs to be done about "The ball's speed is low enough that air resistance is negligible".
 
  • #4
The fact that air resistance is negligible simply states that there is no friction. Dividing a location by a time is meaningless, as is multiplying a location by a time. Your velocity is of the form [tex] (v_x,v_y,v_z)[/tex] where the vertical direction is y. If your only under the influence of gravity, how does [tex]v_y[/tex] change over time?
 
  • #5
LaTeX Code: v_y should decrease over time.
 
  • #6
Yes, and in a gravitational field, you know it decreases by 9.8m/s^2. Thus, you can determine what the [tex]v_y[/tex] velocity is after the time given.

The question might want a different method though, what does the question mean by momentum principle?
 
  • #7
pfinal = pintial + the net force * the change in time.
 
  • #8
Well, that seems a little silly to use that method but ok. Remember, [tex] p = mv [/tex] so [tex] p_f = mv_i + F_{net} * \Delta t[/tex]. You know the force of gravity on an object is [tex] F = mg [/tex] where g = -9.8m/s^2. Thus, you can easily determine the final momentum and thus, final velocity for the y-direction. Of course, the x and z velocities do not change as they are not affected by gravity.
 
  • #9
in the problem, there is no M stated.
 
  • #10
Just say m = M, it will cancel out since the mass does not change
 
  • #11
thanks. i was able to figure out the prob. j/w can you clear me up on graphs of motion? i am a bit confused on how to read them.
 
  • #12
What are you having problems with inregards to them?
 

1. What can I do if I've been stuck on a problem for hours?

Take a break and come back to it with a fresh mind. Sometimes stepping away from a problem can help you see it in a new light.

2. How can I get someone to give me a hint?

Reach out to a colleague, mentor, or friend who may have experience or knowledge in the subject matter. They may be able to provide a useful hint or perspective.

3. Is it better to keep trying to solve the problem on my own or ask for help?

It depends on the situation. Sometimes seeking help from others can save time and provide valuable insights. Other times, pushing through and solving it on your own can be a rewarding learning experience.

4. How can I prevent getting stuck on a problem for hours in the future?

Make sure you fully understand the problem and have all the necessary information before starting. Break the problem down into smaller parts and tackle them one at a time. And don't be afraid to ask for help or take breaks when needed.

5. What are some strategies for solving difficult problems?

Break the problem down into smaller, more manageable parts. Draw diagrams or write out steps to help visualize the problem. Try approaching the problem from a different angle or using a different method. And remember to stay patient and persistent!

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