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What was the angular velocity of the spear when it left his hand?

  1. Sep 24, 2007 #1
    man i dont understand this stuff i have a test on thursday and i found some practice problems can anyone help me solve these( please show work) i need to learn how to do them i have the formulas i think but idont know when and which to use please help soon

    1. In 1993, Bill tossed a ball 201.24m. Suppose bill threw the spear at 35.0 degree angle with respect to the horizontal. What was the angular velocity of the spear when it left his hand?

    2. A softball is through with an angular velocity of 14.5m/s. If you throw the ball at a 35.0 degree angle, calculate the height of the ball at its highest point and how far away the ball is horizontally when it reaches its highest point.

    3. A catapult launches a stone into the air at 50.0 degree angle with an angular velocit of 60.0 m/s. How long will the stoe have been in the air when its velocity is -25.0m/s?

    4.A kangaroo cleared a fence by jumping wit an angular velocity of 8.42m/s at an angle of 55.2 degrees with respect to the ground. If the jump lasted 1.40 s how tall was the fence assuming the kangaroo barely managed to clear it? What was the kangaroos horizontal distplacement for the trip?

    5. You shoot at a basketball goal 3.00 m above your head, making the shot. If the ball is in the air a total of 3.50 seconds and is shot with an angular velocity of 21.0 m/s at what angle did you shoot the ball?

    please and thank you
     
  2. jcsd
  3. Sep 24, 2007 #2
    Welcome to PhysicsForums, gary_shuford.

    In order to receive our help, you must either

    1)Show work or somehow present signs that you have attempted these problems
    2)Specify on which part you are stuck on.

    Remember, we help with your homework, not do your homework.
     
  4. Sep 24, 2007 #3
    yes i understand but i really dont even know the forumals these arent hoemwork there just i need help with them where to start
     
  5. Sep 24, 2007 #4
    First, you must separate the velocity of a projectile into its vertical and its horizontal "component" velocities. The relationship between projectile velocity and its horizontal and vertical components are:

    Vh = V cos(a)

    Vv = V sin(a)

    Where:
    V is the projectile velocity
    Vh is the horizontal velocity component
    Vv is the vertical velocity component
    a is the angle from horizontal.

    As you can see, an angle of zero causes Vh to be the same as V and Vv to be zero. An angle of 90 degrees causes Vv to be the same as V and Vh to be zero.

    Until a projectile hits its target, Vh remains constant (ignoring air resistance). Vv, however, drops every second by 9.81 meters per second. So, for example, Something that has an initial Vv of 15 meters per second will, after two seconds, have a Vv of -4.62 meters per second (the minus means its falling).

    Here is a handy equation describing time, velocity, and acceleration:
    t = V/A
    Where
    t = is how long it takes to reach a velocity of V from an initial velocity of zero, or how long it takes to reach a velocity of zero when the initial velocity was V.
    V = is a velocity
    A = acceleration

    Another handy equation is the one that describes the relationship between time, distance, and acceleration:

    d=A/2 * t^2

    Where:
    d = distance traveled, in meters
    A = acceleration (which is 9.81 meters per second per second for the surface of the Earth
    t = time, in seconds

    The above two equations using acceleration apply only to Vv, because the Earth is pulling downward on the projectile. Remember, Vh doesn't change (if you're ignoring air resistance).

    Now if you have to figure what happens to a projectile that starts with an initial positive Vv (its shot upward) as well as figuring what happens when it comes back down, then you might need to separate the problem into the part where the object goes up to its highest point (the point where Vv == 0) and then solve the part where it falls, as a separate problem. Sometime you won't have to though.

    Here's one more relationship (the simplest one!) among distance, velocity, and time:

    d = v * t
    Where:
    d = distance traveled, in meters
    v = velocity, in meters per second
    t = time, in seconds

    And here's an example problem:

    A canon on a flat plane fires a ball that strikes 300 meters away from the canon. The angle of the canon barrel is 40 degrees. How long does the ball stay in the air. Ignore air resistance.

    As with any problem, you start by writing down everything you know in the form of equations:

    300 = Vh * t
    Vh = V * cos(40o)
    Vv = V * sin(40o)
    t/2 = Vv/9.81

    That last equation describes how long it takes the vertical velocity to come to a stop (at the ball's highest point) plus the time it takes it to fall back down. So the total time is twice the time to rise to the peak.

    Now its just a matter of substitution and a little algebra.
     
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