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Homework Help: Maximum height?

  1. Oct 10, 2005 #1


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    maximum height??

    1) A ball is thrown vertically upward and is caught by the thrower after 2.00s. find a) initial velocity b) the maximum height it reaches?
    a) initial velocity obviously = o, but i have no idea how to find max. height???

    2) the driver of a 600kg sports car, heading directly for a railroad crossing 250m away, applies the brakes i a panic stop. the car is moving at 40m/s, and the brakes can supply a friction force of 1200N. a) how fast is the car moving when it reaches the crossing?

    - i tried finding the change in time for this, thinking i could then find displacment from that, but it ended up as 2.5hours!!
    any help appreciated!!

    3) a metre-stick is found to balance at the 29.7cm mark when placed on a fulcrum, when a 50g mass is attached at the 10cm mark the fulcrum must be moved to the 39.2 cm mark for balance. what is the mass of the meter stick?

    - just plain stuck on this, tried about 3 different ways of attempting the question, but couldnt find one that suited!! again, any help appreciated!!

  2. jcsd
  3. Oct 10, 2005 #2


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    Homework Helper

    Hint on #1: max/min ==> derivative.....
  4. Oct 10, 2005 #3
    Also, your book (or many online resources) may give a formula for the max. height and/or range. I have found that it isn't a good idea to use this. I don't really know why it's put there, but it's definitely better to optimize it using hotvette's suggestion.

  5. Oct 10, 2005 #4


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    the derivative of...???? do i not need to have a range or something to get that?? sorry... im lost
  6. Oct 10, 2005 #5


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    The equation of motion of a projectile is an equation where the vertical travel (y) is a function of angle and time (t) or angle and horizontal distance (x). Thinking in mathematical terms, a function with a max or min can be found by differentiating the equation with respect to the independent variable, setting the equation equal to zero and solving for the independent variable. Example. If [itex]y = 3x^2 + 4x + 1[/itex], how would you find [itex]y_{max}[/itex]?

    What you need is an equation for y as a function of t. Do you have one?
  7. Oct 10, 2005 #6


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    HINT 1: Use energy conservation.

    HINT 2: Use Force = mass times acceleration to find acceleration; use calculated acceleration to determine the speed after travelling a given distance.

    HINT 3: Sum of torques = 0
  8. Oct 10, 2005 #7
    For 1), if the initial velocity was 0 the ball wouldn't go anywhere...

    Distance = V(initial) * T - .5 g T^2
  9. Oct 10, 2005 #8
    in reference to # 1

    think about it like this. since there are no air resistances, you know that the time it takes to go up, say v_uo is equal to the time it takes to come down, say v_down, and that their sum is 2.00s. so from that u know that the half-trip is 1.00s. also, v_0 is not = zero, rather, V_final, the final velocity when the ball reaches its peak is = zero. from that, you should be able to get the initial velocity using v_final = v_initial + accel. x time, where v_final = 0, accel is = gravity, and time = 1.00 s. then with v_final, accel., and the time you can get the max height using delta_y = v_initial x t + (1/2 x accel. x time^2), where v_initial is zero (the velocity at the peak height), and t = 1.00s. :tongue2:
  10. Oct 11, 2005 #9
    Question 1 only

    1. Listen to Skippy.
    2. The motion is symetrical. So it takes half the time to go up and half the time to come down.
    3. Somewhere (book, online) you can find the velocity of any object at the top of its tragectory in freefall motion. It is always true.
    4. Solve only for the down motion (because of hint 2) final and intial velocity are the same. (Don't forget to take only the time it takes to come down.

    With hints 2 and 3 hint #1 becomes trivial and the problem is easy.

    Post a solution with an explanation and i'll help with problem 2. (but lets get the first on before we bite more than we can swallow.)
    Last edited: Oct 11, 2005
  11. Oct 11, 2005 #10


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    seeing as i am only calculating half of the motion, do i use the time as 1.00s or 2.00s??
  12. Oct 11, 2005 #11


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    ok... i think i have got it... i followed what you and skippy said to do, used Distance= v(initial)t-.5gt^2 with v(initial) as 9.8 and time as 1.00s... i came up with 4.9m. this was also true of a different equation that i found online, similar idea and same answer.... YAY!! thanks for that!!!

    any suggestions for 3??? ive tried heaps of ways, but cant find an actual uniform way to do it with a set formula or anything, i think im just making it up!!!

    thanks in advance!!
  13. Oct 18, 2005 #12
    "A metre-stick is found to balance at the 29.7cm mark when placed on a fulcrum. When a 50g mass is attached at the 10cm mark the fulcrum must be moved to the 39.2 cm mark for balance."

    Hey, is there something wrong with the question? When the 50g mass is attached, all the moments about the fulcrum (at 39.2 cm mark) seem to be in one direction only. This does not seem to agree with the Principle of Moments...
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