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Vertically Thrown Rock Part 2

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Brad tosses a rock straight up at 40.0 m/s. What is the rock's velocity after 1.8 seconds? (answer 22.36 m/s)
    What is the velocity of the rock from the question above after 5.2 seconds? (answer -10.96)
    What is the maximum height of the rock from the question above?

    2. Relevant equations
    V=Vo+at
    X=Xo+Vot+1/2at^2
    V^2=Vo^2+2a(X-Xo)

    3. The attempt at a solution
    I tried to find the maximum height for both times, but neither one was the right answer. I did the second equation (X=(40)(1.8)+1/2(-9.8)(3.24) and got 56.124 m and that wasn't right) then I tried the second equation with the second time (X=(40)(5.2)+1/2(-9.8)(27.04) and got the answer to be 75.504), but that wasn't right either. Please tell me what I am doing wrong!
    *hint: the difference between the squares of the initial and final speeds is equal to twice the product of the acceleration due to gravity and the displacement.
     
  2. jcsd
  3. Oct 19, 2011 #2
    how do you know that at max height the time is 1.8s?
     
  4. Oct 19, 2011 #3
    but the max height does not occur at 1.8 seconds or 5.2 seconds.
    you dont know when the max height occurs,no do you care.
    you only need to find the max height, so need distance travelled by rock to point when velocity is 0.
    ie
    u=40 v=0 (at top of motion) s=? a=-g

    now use v^2=u^2+2AS
     
  5. Oct 19, 2011 #4
    I tried V^2=u^2+2AS (0^2)=(40^2)+2(-9.8)s and got 0 but I don't think that's the right answer.
     
  6. Oct 19, 2011 #5
    not sure how you get 0.?

    you have 0=1600-19.6s

    so s=0 certainly does not work
     
  7. Oct 19, 2011 #6
    v[itex]^{2}[/itex] = u[itex]^{2}[/itex] + 2as

    0 = 40[itex]^{2}[/itex] + 2(-9.8)s

    s = 40[itex]^{2}[/itex]/[2(-9.8)]
     
  8. Oct 19, 2011 #7
    Yes, but when you subtract 19.6 from 1600, you get 1580.4s and then don't you have to divide 0 by that number?
     
  9. Oct 19, 2011 #8
    I tried this and it gave me a negative number. (-81.63). It wasn't correct. I am so confused as to what to do!
     
  10. Oct 19, 2011 #9
    i am not going to solve

    0=1600-19.6s

    for you.

    surely you can see if s was negative,-19.6s would be positive so you are adding two positives to get 0 which is absurd.
     
  11. Oct 19, 2011 #10
    Try to be LOGICAL.

    0 = 1600 - 19.6s

    Now add 19.6s to both sides:

    19.6s = 1600 because 19.6s + (19.6s) = 0

    Now divide both sides by 19.6:

    s = 1600/19.6

    s = ...

    Of course you do not have to show my explanation in your solution. I only did that to show you how simple math is when you try your best to be LOGICAL.
     
  12. Oct 19, 2011 #11
    i got it, thank you for your help.
     
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