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Initial velocity formula?

  1. Jun 4, 2005 #1
    Hi all,

    A friend and myself have constructed a very basic potato cannon and we are now looking to find the initial velocity of the potato as it leaves the barrel. I was wondering if there is a formula with which we can calculate this speed. We have found that flight time is the only ‘easy’ variable to measure, although I am sure there are others that we have not thought of yet. For safety reasons we only fire the projectile vertically.

    We came across a very simplified formula which was

    V = g t​

    Where
    v = initial velocity,
    g = speed of gravity,
    t = time taken for the flight of the projectile – from when it leaves the barrel to when it hits the ground​
    I realise that this discounts the effect of air resistance, but is this correct? Or can anyone point me towards a better formula to use.

    We are both in the equivalent of US Grade 10 so our knowledge of this area of mechanics/physics is limited.

    Thanks.
     
  2. jcsd
  3. Jun 4, 2005 #2
    It might be best to start with the full equation:

    v (final) = v (initial) + a x t

    where a is acceleration and t is time.

    You can use g for a (9.8 m/ss or 32.2 ft/ss). This is the acceleration due to gravity at the earths surface. You are probably fine excluding air resistance since a potatoe has a large mass for its surface area and the potatoe will probably not reach terminal velocity. By assuming no air resistance the initial velocity (exiting potatoe gun) would be the same magnitude as the final velocity (hitting ground). Velocity is a vector (magnitude and direction) therefore the initial velocity is positive and the final is negative. Therefore, if you use total flight time you must double the velocity:

    2v = g x t or v = (g x t)/2

    For example, if your total flight time was 8 s then the potatoe velocity would be
    39.2 m/s or 88.2 mi/hr. The equation you showed is fine if you assume that the initial or final velocity are zero. But then you would have to use 1/2 of the total time since you are solving for 1/2 of the trajectory.

    If you can time the first half of the trajectory (up) and the second half (down), seperately, you would get a better idea of the effects of air resistance. I do a lot of projectile labs with my high school physics class and we typically see a 10 to 20 percent increase in time down due to air resistance on the light rockets we use. Therefore, the final velocity, hitting the ground is less than the initial. Our rockets typically go 40m high in about 3.5 s with an exit velocity of 34.1 m/s.
     
  4. Jun 6, 2005 #3
    Thanks for all of your help!

    We achieved a best flight time of 12.2 seconds today giving us an initial velocity of 60.268m/s.

    Is it possible now to work out the total distance travelled by the potato given the initial velocity and taking gravity as the only force acting upon the potato?

    Since the deceleration and acceleration are taken as constant, would this figure just be the average velocity (30.134m/s) times the total time? Thus 367.63m.

    Thanks again.
     
  5. Jun 6, 2005 #4
    Are you familiar with

    [tex] x-x_0 = v_0t + \frac{1}{2}gt^2 [/tex] ?

    If so, then at the very top of the trajectory when the potato is at its max height, it will have no vertical velocity, [itex] v_0 = 0 [/itex], and if we take that as the starting point then we can say [itex] x_0 = 0 [/itex], also. The remaining equation then becomes:

    [tex] x = \frac{1}{2}gt^2 [/itex]

    It takes the same time to get from the cannon to the peak as it does to get back down, thus the time from the peak to the floor is 6.1 seconds. Use this time in the above equation to find how far you fell from.
     
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