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!

Projectile motion: Finding out the time and Dx

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data
    During the opening kickoff of a college football game, the kicker kicks a football with an initial velocity of 27.5m/s at an angle of 41 degrees above horizontal.
    What is the flight time for the ball?
    What is the maximum height
    How far does it travel before hitting the ground?


    2. Relevant equations
    The 4 acceleration equations


    3. The attempt at a solution
    What I know(I'm using components in the by the way)

    Here's a very lame image of what I think the problem looks like
    http://desmond.imageshack.us/Himg580/scaled.php?server=580&filename=forgivemefather.png&res=medium [Broken]
    V1x = 27.5m/s cos41 degrees
    Ay = -9.80
    V1y = 27.5m/s sin41 degrees
    V2y = 0(Please tell me if this is wrong)
    V2x = 27.5m/s cos41 degrees
    Need Dx
    Need Dy
    Need time

    I found Dy already using the V2y^2 = V1y^2 + 2aΔd equation and rearanging it to solve for Δd

    I got 16.6 for the maximum height

    Now I'm trying to find the time, I thought I would be able to do this by rearanging V2 = V1 + aΔt to solve for time
    It comes out like this

    t = v1x - v2x/a
    As you can see this already doesn't work as I'm gonna end up with 0 regardless, and it definately doesn't take 0 seconds for the ball to hit the ground.

    I tried finding out Dx but that wasn't possible(or I'm rearranging wrong)
    What made Dy so easy was that V2y was 0 so rearanging it was simple. But because V1x and V2x are the same rearanging it makes me run into the same problem as when trying to find out time.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 30, 2012 #2

    PeterO

    User Avatar
    Homework Helper

    You can adjust the initial condtions in this applet to see what happens, then se if you can match the figures.
    Perhaps seeing what happens may give you some hints.

    http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/enapplet.html [Broken]

    You can adjust the mass of the ball to see if it makes any difference [try air resistance].
     
    Last edited by a moderator: May 5, 2017
  4. Jan 30, 2012 #3
    I actually already know what the answers are
    It's finding out how to get to those answers is the problem.
     
  5. Jan 30, 2012 #4

    PeterO

    User Avatar
    Homework Helper

    In red above: perfectly good idea - but remember you are only working with the vertical component.
    What is the vertical velocity at maximum height?
    The velocity when landing [coming down] is different to the velocity when starting [going up]
     
    Last edited by a moderator: May 5, 2017
  6. Jan 30, 2012 #5
    Wait
    So there are 2 seperate velocities?
    Doesn't the velocity stay constant throughout the entire thing(at least until it hits the ground)?
    Plus the ball is being thrown at an angle, 2.7m/s should apply for both the X axis and the Y axis.
     
  7. Jan 30, 2012 #6

    PeterO

    User Avatar
    Homework Helper

    if you throw a ball at 2.7 m/s, "almost" vertical, the vertical component is close to 2.7, reduces to zero at maximum height, then returns to almost 2.7 just before hitting the ground.
    The horizontal component, while being very small, remains constant for the duration of the flight.

    If you threw the ball at just 1 degree above the horizontal, the vertical component would be very small, so it would be back at ground level in no time at all.
    The horizontal component would have been almost 2.7, but since the flight was of such short duration, the ball will land "not very far away".

    If you throw a ball at 14.14 m/s at an angle of 45 degrees, both the vertical and horizontal components will be equal, at 10 m/s. [sin45 = cos45]

    That vertical component of 10m/s, if we approximate g to 10ms-2 means it takes 1 second to stop travelling up [reach maximum height], then another second to come down again.
    With a horizontal component of 10m/s, we see the ball will land 20m away [10 m/s for 2 seconds]

    At the start, the speed was 14.14. At the top it was 10, just before landing it is back to 14.14 m/s again.

    Your problem had an angle of 41 degrees, so the vertical and horizontal components are slightly different [the horizontal is slightly larger than the vertical].

    The only quantity common to calculations of vertical and horizontal components is time.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook