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!

Distance Traveled by a Fired Projectile

  1. Oct 13, 2014 #1
    • Warning! Posting template must be used for homework questions.
    Hey everyone!

    I'm new to this site, but I figured I might as well start it off with a bang and go straight into the nitty-gritty. This is a problem from the homework assignment I've been given for tonight, and although I can sorta see that it can be solved, the right series of equations to use is seriously eluding me.

    From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the drawing shows, the bullet puts a hole in a window of another building and hits the wall that faces the window. Using the data in the drawing, determine the distances D and H, which locate the point where the gun was fired. Assume that the bullet does not slow down as it passes through the window.

    Drawing:
    IMG_0844.JPG

    I wasn't too sure about how to proceed with this one. At first I took a look at the drawing and assumed that the place the bullet landed in was at exactly at the bottom of the 0.50 m window such that if a line was drawn between the window and the wall, the bullet would lie exactly on that line. (Not sure if this was unreasonable for me to assume, but I didn't know if there was another way to figure out the bullet's landing location)

    In order for the assumption I made above to be true, then the projectile's velocity had to be such that it would travel 6.9 m horizontally in the time it took to travel -0.5 m vertically. Unfortunately I then realized I didn't know the initial velocity for that, because the projectile was already in flight for an unspecified period of time; as the problem explicitly states that the projectile would not be slowed by impact, the velocity when it impacted the window would not be 0. Thus I could not calculate the velocity from impact against the window to striking the wall.

    I also thought about splitting the figure shown into two shapes: one triangle and one rectangle. The triangle would have a height of H-0.5m and the length would be D, while the rectangle would have a length of D and a height of 0.5m. Then I would use the position equation (Y = Yinitial + Vinitialt + 1/2(g)(t)2) to calculate the time it took to reach the 0.5m window. However, I then realized that I didn't know the distance between the top of the building to that window, so I was once again stumped.

    Maybe there's something obvious here that I'm overlooking, but I'd appreciate a second look by one of you fine gentlemen or ladies.

    Thanks in advance!
     
  2. jcsd
  3. Oct 13, 2014 #2

    Mark44

    Staff: Mentor

    In the time it takes the bullet to travel to the wall it's embedded in, it has to fall H meters. Write an equation that involves t and this distance, and solve for t.

    An assumption that doesn't seem to be stated is that the bullet isn't slowed down due to air resistance. This is probably a reasonable assumption to make, especially in light of the explicit assumption that the bullet doesn't slow down when it goes through the glass window. Write another equation that involves t and the horizontal distance the bullet travels (D + 6.9). From the two equations, you should be able to get t, and from that, D and H.
     
  4. Oct 13, 2014 #3
    I've tried doing what you've said, but it's just ended up confusing me. I made the two equations you recommended, which are as follows:

    -H = 0 + 0(t) + 1/2(-9.8)t2
    -H = -4.9t2
    t = sqrt(H/4.9)

    D + 6.9 = 0 + (340)(t) + 1/2(0)(t2)
    D + 6.9 = 340t
    t = (D + 6.9)/340

    I don't see any common variables other than t here, so I tried make them equal to each other.

    sqrt(H/4.9) = (D + 6.9)/340
    D = 340[sqrt(H/4.9)] - 6.9
    D + 6.9 = 340t
    (340[sqrt(H/4.9)] - 6.9) + 6.9 = 340[sqrt(H/4.9)]
    1 = 1

    I'm not seeing how this can be used to solve for t, so if you could suggest where I should be taking it that would be fantastic!
     
  5. Oct 14, 2014 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi Lexielai! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    After the bullet has entered the window ....

    you can determine the time it takes to cross the room,

    and you know that during this time it falls 0.5m under gravity having started with some as yet undetermined vertical velocity (because it has already been picking up vertical speed during its travel before entering the window).
     
    Last edited by a moderator: May 7, 2017
  6. Oct 14, 2014 #5

    gneill

    User Avatar

    Staff: Mentor

    If I may offer a suggestion (which you are free to ignore!), another approach would be to first convert the trajectory from its parametric form (x and y positions as functions of t) to the form y = f(x).

    Then you have two given points on that trajectory (other than the origin where the bullet is fired from ), so two equations in two unknowns. Don't plug in numbers until the end; do the work symbolically and don't be afraid to lump constants together into new constants when convenient.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Distance Traveled by a Fired Projectile
Loading...