1. Limited time only! Sign up for a free 30min personal 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 with different landing heights

  1. Sep 2, 2011 #1
    1. A handball is projected at an angle of 30 degrees with the horizontal from a 20 meter tower with an initial vertical velocity of 10 ms and an initial horizontal velocity of 20 ms. Find the total time of flight.


    Hi, I understand the concepts of projectiles when the landing height is the same including use of the formulas... however Im really stick when the projectile is launched from a different landing height... I would really appreaciate if someone could walk me thorugh this problem so I can undersatnd it and apply it to the other questions...

    So to me I owuld break down the Horizontal and Vertical velocities 1st

    Vv = 10Sin30 = 5m/s
    Vh = 20Cos30 = 8.66 m/s

    Can someone help me with the way the would approach the problem next.... Or beacuse they gave me seperate Vv and Vh do I not need to break it down further??
     
  2. jcsd
  3. Sep 2, 2011 #2
    Hey bionut,

    First of all, either I misread, or something doesn't make sense in the question. They've already given you the vertical and horizontal velocities (i.e. the components of the velocity). So you wouldn't need to do the process you've done. But - the angle they've given, other from being extra information unnescessary to solve the question doesn't fit in! When you have the x,y components you could calculate the angle alone: arctg(10/20) - and the result isn't 30. Are you sure you wrote the question properly?

    Otherwise -
    The approach here is identical to the approach if the heights are the same.
    The x-axis motion, since there are no forces acting there, is described by:

    x(t) = x0 + v0,xt

    And on the y-axis, since there's a constant force acting (gravity):

    y(t) = y0 + v0,yt + 0.5at2.

    Now choose a coordinate system to work with (decide where the origin is, and what the directions of the x and y axis are - there are several ways to do that). Classically it would be convenient to choose the stone's initial point as the origin, x axis point towards forward motion, and y axis above.
    After doing this, decide what the values of the different variables are - and get your solution. That can be done however only after settling the velocity issue...
     
  4. Sep 2, 2011 #3
    Hi Tomer, thanks for your help...

    the answer is 3.24s,

    But I just need to know how to get there...

    I have seen a few examples with the method you use which is confusing...I think in our Lab class it was a section of the range formula=v^2Sin[itex]\vartheta[/itex]Cos[itex]\vartheta[/itex] + Vcos[itex]\varphi[/itex][itex]\sqrt{}[/itex][itex]\vartheta[/itex](10sintheta))^2 + 2gh/g

    That was used to find time up and time down... Im so confused now ;-(

    But with your formula:
    y(t)= 20 + 10t + 0.5 X 10 t^2
    y(t) = 30t + 5t^2
    5t^2=30t
    t=3[itex]\sqrt{}[/itex]6
    t=1.817... so i assume ive used the formula wrong??... sorryif it is really simple i just dont get it ;-( and you help would be amazing... instead of stiing here for another 2 hours...lol
     
  5. Sep 2, 2011 #4
    I'm glad to help, but I cannot solve it for you - against forum policy (and for good reasons!)

    1. Have you understood the velocity problem? Something there doesn't make sense.
    I'll ignore therefore the angle completely.

    2. Do yourself a favor and don't stick to wacko formulas you half remember from class. There are formulas for everything, but except for newton's laws of kinematics and dynamics, I wouldn't bother remembering anything, aside from a few useful connections.

    You've used "my" formula (which is the one you should use in every projectile problem) - but you've used it wrong.
    Please tell me, where is the origin of your coordinate system, where are the x and y axis pointing to? If you answer that, you'll solve the problem in no time. If you work with axes, you have to define them first.

    3. It's better to spend 2 hours solving a problem, but to understand it deeply, than to get the solution and fail in the exam.

    Good luck :-)
     
  6. Sep 2, 2011 #5
    Hi Tomer, thanks and yes I really would love to understand it for my exam..

    Okay going back to cord.

    So Vx=horizontal velocity (which is constant ) = 20 m/s
    So Vy=vertical velocity (which changes by 9.8m/s) = 10m/s initial which is my Viy?

    Sorry i think im just not used to your method but it looks so much simpler...

    Gping back to the formula... Only the vertical motion affects the flight time is that correct?


    so have I got this right? y(t)= 20 + 10t + 0.5 X 10 t^2
     
  7. Sep 2, 2011 #6
    Good boy.

    You've probably meant that, but it's: vx,0 and vy,0 that we're talking about. About their signs: it's + or -, depending on how you define the y-axis.

    It is, and I wish it were "mine" :-)

    Since there's no wall in the horizon, yes :-)

    Wrong :-) And once again you haven't told me what the direction of the y-axis is.
    Please answer that. It could be pointing either upwards or downwards. Choose one direction, think about the signs of things, and try it again. Notice that by saying for example y0 = 20, you assume that the x-axis is one with the ground, and that the y axis points upwards. Do you understand why? So you've chosen the axis, subconsciously - but then you have to remain consistent. Find the problem in what you've written.
     
  8. Sep 2, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You say that vertical velocity "changes by 9.8m/s" (the units should be m/s^2) so why do you use "10" as the acceleration in your last equation?
     
  9. Sep 2, 2011 #8
    Also true, but more critical is understanding the signs are wrong here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook