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Kinematic Problem verification

  1. Mar 23, 2005 #1
    I'm working on this problem and I've came up with a solution but however I couldn't find anything else to verify the problem so if you could let me know if I did it correctly and help me on the right track what would be super.

    The problem goes as folllow: A player is shooting a ball into a net, he shooting with a 45 degree angle and is 4m away from the net and the net is one meter up. How fast must he throw the ball for it to go in (initial speed).

    I used the equation: D=v1*t+1/2*a*t^2 and split it into X and Y which gives me these equations:
    Y: 1=v1*t+1/2*(-9.8)*t^2 (we allways assume -9.8m/s^2 for gravity)
    X: 4=v1*t+1/2* (0) *t^2
    and solved it gived me 5.11 m/s but since that the X/Y speed we want the longest line of the triangle so I do
    So my answer would be 7.22957m/s

    Is this correct or I am way off?
  2. jcsd
  3. Mar 23, 2005 #2
    [tex] x(t) = V_0cos(45)t [/tex]
    [tex] y(t) = V_0sin(45)t + \frac{gt^2}{2} [/tex]

    You want x(t) to be 4
    and y(t) to be 1

    You should get a system of equations which you solve for V_0. Your answer seems a bit high.
  4. Mar 23, 2005 #3
    The second formula should be :

    [tex] y(t) = V_0sin(45)t - \frac{gt^2}{2} [/tex]

    because the y-axis is pointing upwards and gravity is downwards

  5. Mar 23, 2005 #4
    What I would probably do is calculate your time of flight. You are travelling 4m right and 1m up.

    [tex] T_{flight} = sqrt{\frac{2h}{g}} [/tex] from the equation you provided.

    Then take the time of flight, divide the distance travelled in the x direction by it and you'll get [tex] V_x [/tex]
  6. Mar 23, 2005 #5
    Hey thanks for you help. I used the equations and I got 7.22957m/s aswell so I'm guessing it's safe to assume I got the right answer? Thanks for your help, it's really nice to have resources like this forums to get some help :)

    edit: I just saw your last reply whozum, give me time to check with your equations aswell and I'll get back to you guy with the answer I get.

    edit2: Sorry but what does the h stand for in that equation? I'm guessing the G stand for gravity which in my case would be -9.8m/s
    Last edited: Mar 23, 2005
  7. Mar 23, 2005 #6
    h is the height to travel, h=1m

    For V in the x direction

    [tex] T = sqrt{2h/g} = sqrt{2/9.8} = 0.452 [/tex]

    [tex] d = 4m, t = 0.452s, v = d/t [/tex]

    [tex] v_x = \frac{4m}{0.452s} = 8.85m/s [/tex]

    For V in the y direction

    [tex] y(t) = V_ysin(45)t+\frac{gt^2}{2} [/tex]

    Solving for V_y:

    [tex] V_y = \frac{1-\frac{gt^2}{2}}{sin(45)t} [/tex]

    [tex] V_y = \frac{1-\frac{(-9.8)(0.452)^2}{2}}{sin(45)(0.452)} [/tex]

    [tex] V^2 = V_x^2+V_y^2 [/tex]
    Last edited: Mar 23, 2005
  8. Mar 23, 2005 #7
    Well that's gives me the Sqrt of -.204082 which is a non-real number.
  9. Mar 23, 2005 #8
    Take g to be positive 9.8 instead of -9.8. Look at my work above.
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