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Homework Help: Projectile motion

  1. Sep 12, 2004 #1
    can someone tell me the easiest way to solve projectile motion problems? like how to read and start it? it is extremely hard for me, i just dont get it.
  2. jcsd
  3. Sep 12, 2004 #2
    When you ignore friction and assume the gravitational force is a constant one near the surface of the earth (F=mg) it's quite simple. independent of the mass of the object it falls with an acceleration a=g (Because of Newtons second law F=ma with F=mg -> a=g !).

    In a standard projectile problem you are given the initial conditions like initial speed [itex]v(0)[/itex] and position [itex]x(0)[/itex] and you are to calculate the trajectory ([itex]x(t)[/itex])(the position as a function of time).

    So you have to know how the position of an object is affected by it's acceleration and speed. These relations can be obtained by integrating a=g and using the appropriate initial conditions but I'll give you the results for the constant acceleration g:

    [tex]v(t)=v(0)+ gt[/tex]

    The first equation says the speed will increase (or decrease for negative g) linearly with time with proportionality factor the acceleration, this in addition to the initial velocity v(0). This is intuitively right for acceleration represents the increase in speed.

    The second equation says that under a constant acceleration the position of the object can be obtained by adding the initial position to a few other terms. The second term says your position increases beacause you have a certain velocity, sounds logically right? The second term is there for the velocity increases beacuse of the acceleration.

    Now if you are given g, v(0), x(0) you can find out the trajectory with the above equations in 1 dimension only! This means the motion is in a straight line. An example is an object falling from an air balloon towards the earth with an acceleration a=g=9,8 m/s^2 when thrown upwards with a certain initial velocity v(0) at a certain height x(0). Bu sure to watch the signs of the quantities though (The initial velocity is positive, as is the initial position, but the acceleration is directed downwards so is negative!)

    But ofcourse you would like to find out e.g. what happens if you throw an object vertically to the earths surface while the acceleration (by gravity) acts in the other direction downwards. Then you would have to use vectors. Instead of [itex]x, v, g[/itex] you would use [itex]\vec{x}, \vec{v}, \vec{g}[/itex] and the given equations will still hold. But I'm not sure you are too familiar with vectors (three numbers describing the quantities in the three spatial directions) so I'll try a different approach (basically the same).

    When you will try to solve a problem concerning motion you will have to pay attention to the directions involved. Basically what you have to do is use the equations above for every direction involved seperately. An example:

    When you fire an object in an horizontal direction (let's call this the x-direction) it will maintain that same speed in time. so [itex]x(t)=x(0)+v(0)t[/itex] (g=0!). But in the vertical direction (let's call this the z-direction) there is an acceleration beacuse of gravity equal to g=9,8m/s^2 an initial height (let's call this h) but no initial speed, so: [itex]z(t)=z(0)-.5gt^2[/itex] (Note the minus sign!). This gives you the trajectory of the projectile.

    If you woul like to know what the velocity is as a function of time you use the other formula. Note that you get 2 velocities, one velocity in the x- and one in the z- direction, if you want the total speed you calculate [itex]\sqrt{v_x^2+v_y^2}[/itex])

    If you would like to know at what time the projectile lands you do the following. When it lands its height (z-coordinate) is 0. So you fill in z(t)=z(0)-.5gt^2=0[/itex] and solve for t. This yields [tex]t= \sqrt{ 2z(0)/g } [/itex]

    Another example:
    If the projectile is launched under a certain angle (let's call this one w) with the horizontal. The first thing you would do is determine the initial conditions. The initial height, velocity and angle are given, but you would like to know what the initial velocity is in the x-direction and the initial velocity in the z-direction. What you do is you draw a graph representing the velocity as an arrow (a vector!) under a certain angle with the horizontal. By using goniometry you can find out that the initial velocity in the x-direction (horizaontal) is v(0)cos(w) and the intial velocity in the the z-direction (vertical) is v(0)sin(w). Now you can use the above formula to find out everything you want

    A general scheme to work out the problem might be:

    -draw a picture: this helps you visualize the problem. Don't forget to put in as much data as you can. The directions along wich you are going to work, angles, initial velocities, etc.
    -Work out the initial conditions: the initial positions and velocities in the different directions.
    -Solve the original problem using the above formulas (one for each direction!)

    Good Luck!!
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