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Thread: Physic two dementional motion

  1. #1
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    Default Physic two dementional motion

    Ok so I've tried to do these problem multiple times and every time I get the wrong answer

    1. A basketball player makes a jump shot giving the ball a velocity of 10.0 m/s at 60(deg)N of E. Assume the ball is released at the same height as the basket. How far away from the basket was he when he made his shot
    Book answer: 8.85 m
    My answer: 4.4 m

    My work
    viy = sin(60)10 = 8.67
    Vx = cos(60)10 = 5
    a = -9.8

    The rest I am unsure of, but I get 4.4m I set Vf equal to 0 and solved.

    2. A daredevil is shot out of a cannon at 45(deg) N of E with an initial velocity of 25m/s. A net is position at a horizontal distance of 50.0m. At what height above the cannon should the net be positioned in order to catch the daredevil?
    Book answer: 10.9m
    My answer: 11.08m

    My work
    Vx = cos(45)25 = 17.68
    Viy = sin(45)25 = 17.68
    Change in X = 50
    a = -9.8
    t = 2.8 sec (found this on out using a formula) 17.68 = 50/t

    3. (I just cannot figure this one out!)
    You throw a softball at an unknown angle (N of E). It's in the air for 4 seconds total and travels 50m horizontally, calculate the velocity of the ball in MPH and the angle it was thrown at. Assume it landed at the same height.



    Any help is greatly appreciated!

  2. #2
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    First one: v_fy is not going to be zero. The y-velocity at the top of the arc is zero. Calculate the time it will take to go from 8.67 to 0 m/s, then multiply it by two, to get the time for the whole flight (dv = a*dt). Then, solving for the x-distance is trivial, since there is no acceleration in x (d = v*t).

    2nd one: Your answer looks pretty good. Try to be consistent with the number of decimal places you use in your calculations vs. how many the book is using.

    3rd: Similar to the first one. You know it's going to move in a parabola, so the time to reach the peak of its flight will be half the total time, so 2 s. You know the y-speed at the top of the arc has to be 0, so you have
    delta v = 0-viy = -viy
    t = 4/2 = 2 s
    and a = -g = -9.8 m/s^2
    Now use dv = a*dt to find the initial y-velocity.
    Lastly, since there is no acceleration in x, the initial x speed is just the average, 50 m/4 s = 12.5 m/s

    Now you have
    vix = vi*cos(a)
    viy = vi*sin(a),
    Which is two equations in two unknowns and solvable.

  3. #3
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    You prob have the answer by now, but. Sorry for gravedigging.
    (Didn't read your whole post)
    You have to break it into 2 parts,
    1)Finding out how long the ball would take to travel to the highest point and then back down to the point it was released. Since it is symmetrical you can just find the answer going up and then multiply by 2.
    2)You use this time and the horizontal velocity (acceleration 0) to find the distance X.

    D= T X V

    Remember to always break down your vectors into components. Use the X component. For the first part and the Y for the horizontal.
    Tip: Not always vi(cos)/ vi(sin) for the components!! Take a look at the angle!

    Tip2: IF the motion is symmetrical up and down AND no added acceleration you can multiply time of "flight "by 2 to find the total time.

    Didn't read your work but your problem might be this last tip where you only found the "up part." Multiply by 2 Fix rounding issues and you get 8.85



    For any 2D projectile motions you use basically the same 2 steps just foward/backwards... and sideways? :P

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