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Motion of projectiles

  1. Jun 23, 2004 #1
    I keep getting the wrong answer! For this question, I'm getting 4.70 m...can someone tell me what i'm doing wrong?

    A ball is thrown from a point 1.2m above the ground. The initial velocity is 19.2m/s at an angle of 30 degrees above the horizontal. Find the maximum height of the ball above the ground.

    I'm taking the (initial velocity multiplied by sin 30)^2 / 2*9.8 which is 4.70...but it's not correct. What am i doing wrong?
     
  2. jcsd
  3. Jun 23, 2004 #2
    You can check these types of problems using conservation of mechanical energy. Let h = 1.2 meters, v = 19.2 m/s, H be the maximum height, u the velocity at this height and m the mass of the ball. Then we have
    [tex]mgh + mv^2/2 = mgH + mu^2/2 \rightarrow 2gh + v^2 = 2gH + u^2[/tex]​
    Now, u is just the horizontal component of the initial velocity which is [itex]v\cos{\theta}[/itex] where [itex]\theta[/itex] is 30 degrees in this case. Solving for H gives
    [tex]H = \frac{2gh + v^2 - u^2}{2g}[/tex].​
    Plug in the values and see what you get.
     
  4. Jun 23, 2004 #3

    Doc Al

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    Staff: Mentor

    That equation, based upon v2 = 2ax, will find the maximum height above the starting point. You are forgetting that it didn't start on the ground.
     
  5. Jun 23, 2004 #4
    Now i see. I needed to take the 4.70m and add the 1.2 meters in order to get the maximum height. Thank you!!
     
  6. Jun 27, 2004 #5
    Hi...I believe your problem is solved but sorry for the long post:

    Just to add to this, you are normally better off using the more general relationship,

    [tex]
    y = y_{0} + (V_{0}sin\theta)t - \frac{1}{2}gt^{2}
    [/tex]

    where y_{0} is the initial y-coordinate of the body (it does not necessarily start from y = 0), V_{0} is the magnitude of the projection velocity, theta is the angle made by the direction of projection with the horizontal and of course t and g are time and acceleration due to gravity.

    As you have to find the maximum height, which would be given by

    [tex]
    y_{max} = \frac{V_{0}sin^{2}\theta}{g}
    [/tex]

    if the projectile were projected from the origin (x = 0, y = 0).

    Now since we're projecting it from some a point whose y coordinate is y_{0}, the relation for y_{max} changes as if the axes have been translated to the point of projection. So, now the correct relation is

    [tex]
    y_{max} = y_{0} + \frac{V_{0}sin^{2}\theta}{g}
    [/tex]

    This is the equation you need for your problem. You can verify it rigourously by using the first equation above and taking its time derivative. This yields,

    [tex]
    \frac{dy}{dt} = V_{0}sin\theta - gt
    [/tex]

    Setting this equal to zero you get the time at which the maximum height is reached,

    [tex]
    t_{ymax} = \frac{V_{0}sin\theta}{g}
    [/tex]

    Now put this value of t_{ymax} into the first equation and you will get the abovementioned formula for y_{max} in the general case where the particle/projectile is projected from a point whose y-coordinate is y_{0}. In fact, if you have studied parabolas in mathematics, you will be able to associate the characteristics of parabolas with the trajectory of projectiles (which is parabolic as you already know). This helps in quickly transforming the equations.

    Finally, you should keep in mind the following equations for the general case of projection from the point (x0, y0):

    [tex]
    x = x_{0} + (V_{0}cos\theta)t
    [/tex]
    [tex]
    y = y_{0} + (V_{0}sin\theta)t - \frac{1}{2}gt^{2}
    [/tex]
    [tex]
    V_{x} = V_{0}cos\theta
    [/tex]
    [tex]
    V_{y} = (V_{0}sin\theta)t - gt
    [/tex]

    and the equation of trajectory:

    [tex]
    y = y_{0} + (x-x_{0})tan\theta - \frac{1}{2}\frac{g(x-x_{0})^{2}}{V_{0}^2}sec^2\theta
    [/tex]

    Hope that helps...

    Cheers
    Vivek
     
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