1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation for the max height reached by a projectile

  1. Apr 3, 2007 #1
    A projectile of mass 20.2 kg is fired at an angle of 65.0 above the horizontal and with a speed of 84.0 m/s. At the highest point of its trajectory the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. How far from the point of firing does the other fragment strike if the terrain is level?

    I know this has something to do with p=m(v_center of mass), but I'm not really sure how to deterine the acceleration of the center of mass. It is constant up until the explosion, but then the explosion stops one half of the project dead in its tracks where it falls straight down, whereas the other half of the projectile is accelerated with an equal but opposite direction as the first half of the projectile. I do not know how to quantify that.

    I was able to get answer with the following equations, though I am unsure as to whether I used a correct approach.

    The equation for the max height reached by a projectile is :

    m.h. = [V_0^2(sin^2(theta)]/2g
    m.h. = [(84.0 m/s)^2(sin^2(65))]/19.60 m/s^2
    m.h. = 295.7 m

    I deduced this distance along the ground using (R = the hypotenuse):

    Rsin(theta) = height
    R = height/sin(theta)
    R = 295.7m/sin(65)
    R = 326.27m

    Rcos(theta) = distance along ground

    This is the point to which I calculated the half of the projectile that falls vertically down from the max height to land. To find the position that the center of mass lies along the ground, I used the equation for max range of a projectile.

    m.r. = [(V_0)^2(sin(theta))]/g
    m.r. = [(84.0 m/s)^2(sin(65))]/9.8 m/s^2
    m.r. = 652.54m

    This is where I calculated the center of mass of the two halves of the projectile to be. To get the distance of the other half of the particle (that was accelerated forward), I subtracted the distance of the first half by the position of the center of mass, then added this value to the center of mass.

    652.54m - 137.89m = 514.65m
    514.65m + 652.54m = 1167.19

    The distance the particles lie from each other is the difference of the two distances:

    1167.19m - 137.89m = 1029.3m

    Is this a correct approach? If not, then how do I figure out the acceleration after the explosion?
  2. jcsd
  3. Apr 3, 2007 #2
    What a minute, you used the range equation for height (don't just put in equations you learn without having a reason to).

    You are correct that this deals with momentum, and one of the most fundamental physical laws - momentum is ALWAYS conserved. What do you suppose the initial momentum of the projectile (when at the top of the trajectory) is? If one of the fragments merely falls with no extra forces, what do you think that momentum will be? What does that mean for the momentum of the other?

    Essentially, this is just a more involved vector projectile problem.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Equation for the max height reached by a projectile