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Projectile:Launching and landing at different heights

  1. Nov 22, 2007 #1
    1. The problem statement, all variables and given/known data

    1.If a shot is put an angle of 41 degrees relative to the horizontal with a velocity of 36 ft/s in the direction of the put, what will be the upward (vertical) velocity at the instant of release? What will be the forward (horizontal) velocity?
    How high (above the point of release) will the shot go? What is the time it takes the shot to reach its maximum height?

    2.If the shot in the problem above is released from a height of 6 ft and later lands on the ground (height = 0.0 ft),
    what was the total time of flight? How far did the shot travel horizontally?


    2. Relevant equations

    vx=vcos41 = 27.16 ft/s
    vy = vsin41 = 23.6 ft/s
    dy= (vfy^2-viy^2)/2a) => (0-23.61^2)/[2*(-32)] = 8.71m
    vfy = viy + at => 0 = 23.6 + (-32)t => t = 0.7375s

    3. The attempt at a solution

    for the second part im not sure how to proceed do i use the following equation:
    dy= viy^2 + 1/2at^2
    6 = 23.6t -16t^t
    olving for the quadratic equation i get two value for t
    t1 = 1.14848205556934
    t2 = 0.32651794443066

    By common sense when the height was zero and if i was asked to calculate the total time it would have been 0.7375*2 = 1.475 s. So in part 2 the time should be greater than 1.475s since the height is included so can anyone put me on the correct track please ?
     
  2. jcsd
  3. Nov 22, 2007 #2
    I wrote the quetion and where ive reached in a word file heres a print screen :


    [​IMG]
     
  4. Nov 22, 2007 #3

    rl.bhat

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    6 = 23.6t -16t^t
    Here displacement and acceleration are in the same direction and initial velocity is in the opposite direction. So the equation becoms 6 = -23.6t + 16t^2. Now try
     
  5. Nov 22, 2007 #4
  6. Nov 22, 2007 #5

    rl.bhat

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    You won't get imaginary values. Use this equation: t = {-b + or -[b^2 - 4ac]^1/2}/2a
     
  7. Nov 22, 2007 #6
    to be sure that we are on the same page i can see that youve used the same equation i did.

    d= viyt + 1/2at^2

    6 = 23.6t -16t^2

    Rearranging the equation =>

    16t^2 -23.6t +6 = 0


    can u explain why you have it to -6 and not +6 ?
     
  8. Nov 22, 2007 #7
    Trying both equations;

    16t^2 -23.6t +6 = 0

    and

    16t^2 -23.6t -6 = 0

    will still give me time < 0.7375*2

    can anyone gimme a hint ?
     
  9. Nov 22, 2007 #8

    rl.bhat

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    16t^2 -23.6t -6 = 0

    t = {-(-23.6) + or - [23.6^2 - (-4*16*6)]^1/2}/2*16 solve this and take + value because the time cannot be negative.
     
  10. Nov 22, 2007 #9
    16t^2 -23.6t -6 = 0

    a=16
    b=-23.6
    c=-6

    Delta= b^2-4ac
    Delta =(23.6)^2-(4*16*-6)
    Delta = 940.96
    sqrt(delta) = 30.67

    t1= [-b+sqrt(delta)]/2a
    t1= (-23.6+30.67)/(16*2)
    t1=0.22s

    t2= -b -sqrt(delta)2a
    t2= (-23.6-30.67)/(16*2)
    t2 = 1.695s

    using common sense i should take the second one.
     
  11. Nov 22, 2007 #10


    Can you explain to me how did you relate

    Displacement , Acceleration and Initial velcoity together ? to me i wouldnt have solved it if it came up in a test. So what is the difference between part 2 and part 1 ; why did you take a = 32 and not -32 as i did in part 1.
     
  12. Nov 22, 2007 #11

    rl.bhat

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    t1= [-(-b)+sqrt(delta)]/2a
    t1= (+23.6+30.67)/(16*2)
    t1=1.695s

    t2= -b -sqrt(delta)2a
    t2= (23.6-30.67)/(16*2)
    t2 = -.0.22 it is not possible.
    In the first part initial velocity and displacement are upwards and acceleration in the downwards. Out of three values mejority should be positive. In the second case displacement and acceleration are downwards and initial velocity is upwards. Hence the signs are applied.
     
  13. Nov 22, 2007 #12
    So this concept applies for all problems that have different launching and landing wheras a is considered -32 or -9.8 for problems that have the same launching and landing(height =0)

    Question in part 1 is to find the upward so a = -32.
    in the 2nd part to find the total time and distance meaning a is pointing downward = + 32

    I drew this quickly:

    [​IMG]
     
    Last edited: Nov 22, 2007
  14. Nov 22, 2007 #13

    rl.bhat

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    Yes. Nice drawing. Keep it up.
     
  15. Nov 22, 2007 #14
    Thanks man. you were more than helpful to me.
     
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