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Need help w/ velocity problem

  1. Oct 31, 2004 #1
    Any help will be appreciated! :rofl:

    An object is fired upward at a speed of 3.000 m/s at t= 0.0000s. It reaches its maximum height at time t. What is its velocity at the maximum height?
    Assuming that g= -9.81 m/s/s, calculate the maximum height reached in meters: Calculate the time, t, at maximum height:
  2. jcsd
  3. Oct 31, 2004 #2


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    This should go in the homwork help section. Can you show us what you've done so far to tackle it? Also, the first question you should be able to answer without doing anything. Think about it; it has reached its maximum height. There must be some reason why it does not go up any farther than that. What does this say about the velocity at that instant?

    The second part simply involves applying one of the kinematic formulas for objects moving under constant acceleration.
  4. Oct 31, 2004 #3
    well i started it from looking at my notes, this is a hs physic problem, bu ti think u should find the average velocity and and t. Im not exactly sure, physics confuses me!
  5. Oct 31, 2004 #4
    V(F)2 = V(I)2 + 2AD OR D= v(I)T + 1/2AT2
    not exactly sure scatter-brained!!
  6. Oct 31, 2004 #5


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    Oh, ok...

    Well, for the first part, what I was hinting at was that it starts off with an upward velocity, but there is a downward force slowing...it...down...to the point that eventually it ceases to move upward. At that instant, it's upward velocity is completely gone i.e. it has stopped. So the velocity at maximum height is zero. However, the gravitational force is still acting on it, accelerating it downward. So it's velocity will begin to increase in the opposite direction (downward). Is the situation intuitive now?

    As for the second part. The key strategy for these kinematics problems is to list what information you have been given. What info have you been given? Read the problem:


    vi = 3.000 m/s

    vf = 0

    a = g = -9.81m/s2

    Now, have you been given the time required for the object to reach max height? No. In fact, you are asked to solve for it! So, you cannot use the second formula, can you? However, do you have enough info to use the first formula to solve for d?

    And once you have d, do you have enough info use the 2nd formula to solve for t?
  7. Oct 31, 2004 #6
    Try equating the initial kinetic energy with its potential energy at the maximum height... 0.5*m*v^2 = m*g*h ...Am I giving too much away, here? ;)
  8. Nov 1, 2004 #7
    ok so i used 1st formula got .4587 for d? doesnt seem right! then put in givens for second equation and stuck! I dont know if im solving this correctly!

    0.4587 = 3.000t -4.905(t)squared -where do i go from there if thats even
  9. Nov 1, 2004 #8
    "ok so i used 1st formula got .4587 for d? doesnt seem right!"

    It is. The method I suggested gets the same answer: 0.5*m*v^2 = m*g*h, (0.5*9)/9.81 = 0.459 m
  10. Nov 1, 2004 #9
    "0.4587 = 3.000t -4.905(t)squared -where do i go from there if thats even

    It's correct. Next...Hint: x = [-b +- sqrt(b^2 - 4*a*c)]/(2*a)
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