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PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4s

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    An object fired with a velocity of Vo covers a verticle distance of 20.0m and a horizontal distance of 25.0m in the 4th second of its motion. Determine the following:
    a) Its velocity of projection, Vo
    (the rest im not going to worry about cos if i get part a) started then i should be fine i just dont know what im doing wrong)


    2. Relevant equations
    ucosθ = x
    usinθ = y
    v = u + at2
    s = ut + 1/2at2
    v2 = u2 + 2as
    v = s/t


    3. The attempt at a solution

    ok so i've tried many a times but i think i have a misconception or not seeing what is happening etc.
    i know the answer as it is given which is Vo = 59.8 m/s 65.3o (angle between ground and vector)

    so it takes 4 seconds for the projectile to be displaced horizontally by 25 m
    so by v=s/t
    v=25/4
    v=6.25
    which v is also known as x
    so i have x=6.25 m/s

    the verticle velocity i found by many ways or tried to but heres one way
    s=ut + 1/2at2
    s = 20
    a = -9.8
    t = 4
    20 = 4u + 1/2.(9.8).42
    98.4 = 4u
    u = 24.6 m/s
    and in this case u = y
    so y=24.6

    by pythagoras
    Vo2 = y2 + x2
    Vo2 = 24.62 + 6.252
    Vo = 25.2815
    clearly not the answer but i proceeded to find the angle
    y = usinθ
    θ = 14.255
    angle between ground and vector = 90 - 14.255
    angle = 75.745 degrees
    which is clearly not the case

    if you see where ive gone wrong please do tell :)
    ive tried many many wasy of getting the projected velocity and nothing seemed to work, ive tried 4 times and that takes a long time
     
  2. jcsd
  3. Nov 5, 2009 #2
    Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

    sorry i used all caps for the topic, diddnt read the rules properly till it was too late, i do not wish to offend anyone soz
     
  4. Nov 5, 2009 #3
    Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

    I don't know, but I did the exact same thing as you.
     
  5. Nov 5, 2009 #4
    Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

    me too. theres seems to be nothing wrong with your method. i cant think of another way of doing it
     
  6. Nov 6, 2009 #5
    Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

    yeah i found out what was wrong
    it says in the 4th second meaning between t=3 and t=4 so it takes 1 second for the ball to reach 20m high and 25m horizontally haha damn english, ill post the answers soon
     
  7. Nov 6, 2009 #6
    Re: PROJECTILE MOTION - Given values are Horizontal & Verticle distance when time = 4

    1. The problem statement, all variables and given/known data
    An object fired with a velocity of Vo covers a verticle distance of 20.0m and a horizontal distance of 25.0m in the 4th second of its motion.(Meaning second 3 to second 4) Determine the following:
    a) Its velocity of projection, Vo
    (the rest im not going to worry about cos if i get part a) started then i should be fine i just dont know what im doing wrong)

    2. Relevant equations
    ucosθ = x
    usinθ = y
    v = u + at
    s = ut + 1/2at2
    v2 = u2 + 2as
    v = s/t


    3. The Solution

    Vertical Velocity

    s = ut + 1/2at2

    when t=3
    s1 = 3u + ½ X (-9.8) X 32

    when t=4
    s2 = 4u + ½ X (-9.8) X 42

    s2 – s1 = 20m

    s2 – s1 = (4u + ½ X (-9.8) X 42) – (3u + ½ X (-9.8) X 32)

    20 = 4u – 78.4 – 3u + 44.1
    20 = u – 34.3
    u = 54.3 m/s
    let u = y
    y = 54.3 m/s

    Horizontal Velocity

    v = s/t
    v = 25/1
    v = 25 m/s
    let v =x
    x = 25 m/s

    Vo

    Vo2 = y2 + x2
    Vo2 = 54.32 + 252
    Vo = 59.779 m/s

    tanθ = 54.3/25
    θ= 65.278o

    Therefore
    Vo = 59.779 m/s 65.278o (angle of elevation)
     
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