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Free Fall and a Projectile motion question

  1. Dec 7, 2016 #1
    • Member advised to place separate questions in separate threads (One question per thread)
    1. The problem statement, all variables and given/known data
    1. A diver jumps from a height of 10m leaving with a velocity of 8m/s find the time for which the diver is in the air and the speed at which the diver enters the water?

    2. A ball is launched from point A on a horizontal plane with an angle of 55*. It passes two walls both 2m high. Wall one is 2m from point A. Find the speed of the ball and the Wall two distance from point A

    The attempt at a solution

    1. Vy= 8m/s dy = 10m ay=-9.8m/s2

    Vfy2 = Viy2 + 2ayd

    Vy=√(8m/s)2 + 2(-9.8m/s2)(-10m)
    Vy = 16.1 m/s

    -10 m = 8 m/s t + 1/2 (-9.8m/s2) t2
    0 = -4.9 m/s2 t2 + 8m/s t + 10m

    t = 2.45 s or - 0.82 s

    2. Vx = VicosΘ and Vy = VisinΘ h =2m and horizontal distance = 2m

    dy= Viyt + 1/2 at2 dx = Vixt + 1/2at2
    since horizontal acceleration = 0 ; dx = Vixt
    dy= Viyt + 1/2 at2 dx = Vixt ; 2 m = VicosΘ t
    2 = VisinΘ t - 1/2 (9.8m/s2) t 2
    2 = VisinΘ (2m /VicosΘ) - 1/2 (9.8m/s2)(2m /VicosΘ)2
    2 = 2m tan Θ - 4.9m/s2 (12.15 m 2 /Vi2 )
    2 = 2.85 m - 59.535 / Vi2
    Vi = √70.04
    Vi = 8.36 m/s

    Im stuck on how to get the distance assuming this is correct. If I use the max range formula

    x = Vi2 sin2Θ / g
    x = (8.36m/s)2 sin 2(55) / 9.8 m/s2
    x = 6.69 m

    If I take half this distance ( 6.69m)(.5) = 3.34 m
    I know that it is reaching a height of 2 m twice, and given that the parabola is symmetrical I know that on the opposite side it is also 2m away from Wall 2.
    3.34 m - 2 m = 1.34 m , which the remaining distance between the wall to the half the max range.

    therefore 1.34m x 2 = 2.68m which is the total distance between Wall 1 and Wall 2

    So to find the distance from Wall 2 to Point A just add the total = 2.68m + 2 m = 4.68m

     
    Last edited: Dec 7, 2016
  2. jcsd
  3. Dec 7, 2016 #2

    PeroK

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    Does ##21.6s## not seem a long time to be in the air?
     
  4. Dec 7, 2016 #3
    Youre right... my math was off
    its t = 2.45s or -0.82

    assuming everything else is correct
     
  5. Dec 7, 2016 #4

    PeroK

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    You have to make your mind up about whether it's ##2.45s## or ##-0.82s##. Do you understand why there are two solutions?

    In question 2, are you supposed to assume that the ball just clears each wall?
     
  6. Dec 7, 2016 #5
    The time is t = 2.45 s
    For #2 yes you assume it clears each wall
     
  7. Dec 7, 2016 #6

    PeroK

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    I really like your approach to question 2. Using the range (or you could have used the distance to the highest point) and then the symmetry of the parabola was very neat.

    One suggestion is to try to use algebra a little more and not be so quick to plug in the numbers.
     
  8. Dec 7, 2016 #7
    Thanks ; But did I solve the 2 problems correctly?
     
  9. Dec 7, 2016 #8

    PeroK

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    Yes, of course!
     
  10. Dec 7, 2016 #9

    PeroK

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    PS I think the simpler approach was to find the two times that ##y = 2m## and the use the x-velocity to get the distance to the second wall.
     
  11. Dec 7, 2016 #10
    Thanks for the help , I drew a diagram and I see what you mean.
     
  12. Dec 7, 2016 #11

    PeroK

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    Here's a thing. I reckon that if ##h## is the height of the walls, ##a## is the distance to the first wall and ##b## the distance to the second wall, then:

    ##b = \frac{ah}{a\tan \theta - h}##

    In this case ##a = h = 2m## so:

    ##b = \frac{2}{\tan \theta - 1}m##

    Is that interesting?
     
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