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!

Steps to solve Projectile problem

  1. Aug 26, 2004 #1
    Hello- I have a projectile problem that I would like some advice on to approach.
    A rocket is launched at 53 degrees with 75 m/s, and 25 m/s squared for twenty five seconds.
    The question is how high does it go, total time in flight, and finally the horizontal range. Book answer 24,000 m, 152 sec, 78,000 m.

    I approached the problem by breaking it up into two phases- rocket motor on for 25 seconds, second phase motor off and start of return trip.

    The problem seem to center around the acceleration factor and the what happens to projectile when the motor is off.
    I was close on my answers but my work was not a thing of beauty, I am doing my work with a high school text and need to stay at that level before moving on to higher study material.
    Thanks in advance for your time and trouble
  2. jcsd
  3. Aug 26, 2004 #2
    I agree with splitting it up to two parts.
    So first part convert the velocity to X and Y vectors:
    Vx = 75Cos53
    Vy = 75Sin53
    obviously the acceleration is in the same direction, and we can assume the rocket doesnt rotate in the first 25 seconds (assume its a point particle)

    Ax = 25Cos53
    Ay = 25Sin53

    Sum your forces:

    Sum(Fx = m*Ax)
    Sum(Fy = m*Ay - m*g)
    Since the acceleration is velocity and height independent we can say the
    Sum(Fy = m*(Ay-g))

    Wait, before I go farther I must ask what is meant by it "has an acceleration". Does that mean it has a force at that acceleration or that its total magnitude of acceleration vectors is 25, including the force due to gravity.

    I guess I'll just assume its actually accelerating at that speed. So forget above work.

    A = dv/dt
    Adt = dv
    a(t-to) = (V-Vo)
    For x : 25Cos53 * 25 + 75Cos53 = Vfx
    For y : 25Sin53 * 25 + 75Sin53 = Vfy
    Vfx = 421.27
    Vfy = 559.04

    Adt = dv
    At = V-vo
    At + vo = dx/dt
    At+vo dt = dx // integrate
    1/2 At^2 + vo*t = x
    So 1/2 (25Cos53)(25^2) + 75Cos53*25 = x
    1/2 (25Sin53)(25^2) + 75Sin53*25 = y
    Xf = 5830.08
    Yf = 7736.78

    Then part 2 only accel is downward so Vfx2 = Vfx = 421.27
    We need to find T. Id actually split it up to another 2 parts, for coasting to peak then freefall down.
    Vfx = 421.27
    Vfy = 559.04
    Xf = 5830.08
    Yf = 7736.78 // Just for reference
    at = (V-Vo)
    so (0 - Vfy) = -9.81m/s * t
    time = 56.99 Seconds for coast
    For y : (1/2) (--159369.81) (57)^2 + + 559.04*57 = -15936.3 + 31865.28 = 15928.98 meters is height from coasting.
    Total height traveled : 15928.98 + 7736.78 = 23665.76m
    then Freefall from that height = x = .5 a t^2 = 0.5 * 9.81 * t^2 = 23665.76
    t = 69.46 seconds

    So TOTAL time is 69.46 + 56.99 + 25 = 151.4509 seconds ( a little rounding error)

    Then Total x would be Xf from part one + Vxf*t = 5830.08+421.27*(69.46+56.99) = 59100m

    Their answer is wrong (78,000). They included the original 25 seconds in their final multipliction of time*Vxf.
    Last edited: Aug 26, 2004
  4. Aug 27, 2004 #3
    Actually there are just a few basic kinematic relationships for projectile motion and it's normally a good idea to remember them (with time you know them automatically) instead of starting out from first principles in every problem (this problem is a sort of exception from this "rule" but you can still breakup the motion and use acceleration for the first 25 s and free projectile motion subsequently). This is particularly important in an exam, but if you're working you might as well do things from the ground up.
  5. Aug 27, 2004 #4


    User Avatar
    Science Advisor
    Homework Helper

    I disagree completely here. You should do it from the start constantly, so that you never forget for next year. Also, most physics exams tend to want work and not just answers.

    If you do the work enough it gets to the point where it only takes 30 seconds... especially if you have a good calculator.
  6. Aug 27, 2004 #5
    I know I could have done the second half as one, due to time being a squared term (parabolic), but I got used to doing it this way because ive had problem for total path traveled in the Y direction(up and down sum) and its easier to do it by just splitting it up.

    And yes, I like starting from the basics because it constantly reminds you about what everything MEANS, not just formulae to solve with.
  7. Aug 28, 2004 #6
    If you guys thought I am asking you (or hinting) to memorize formulae, you misunderstood me. I just said that you should know what you're doing. Of course, if you start integrating from the acceleration equation and substitute constants, you will get the right answer but you are less likely to get a feel of the situation (I also made an exception about this problem, but I think neither of you read my post carefully...nevertheless).

    Just to be sure, I always prefer starting from first principles, but I don't merely use calculus. I try and use basic stuff like writing what I have and what I need to know. Next I try and determine what I need to do now to reach from what I know to what I don't. At that stage I may as well do exactly what you have done. But that depends on the problem. So I meant don't blindly resort to a long method in an exam. I said when you're studying, its always englightening to do things from more than one method.

    Enjoy physics....


    EDIT: Back here, we're discouraged from using a calculator for the simple reason that it isn't allowed on tests, for your info. So we do EVERYTHING long hand (which is one reason whY I emphasized on a smart method on an exam...if you have a calculato,r you are at liberty to do whatever you want :approve:)
    Last edited: Aug 28, 2004
  8. Aug 28, 2004 #7


    User Avatar

    Are you from India maverick? All of the students I know from India are pretty good with doing calculations in their head, since they can't use calculators. I know of one guy in particular who had memorized pages of log tables :yuck: !
  9. Aug 30, 2004 #8
    Hello Gza

    Yes I am from India. And fortunately I haven't heard of anyone out here who has memorized log tables. I would never be induced to do that. :tongue: Some school boards in India do allow the use of calculators in mathematics, physics and chemistry exams held in the final year of school (we call it Class 12), but all major engineering/medical/etc entrance examinations, olympiads, tests and other exams do not allow the use of calculators. In fact, nowadays, even log tables are disallowed! So I guess it really does pay to memorize log tables...heheh... :biggrin:

  10. Aug 30, 2004 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Memorize log tables...naaaah !! Though from the logs of 2,3,5,7 you can get pretty close to all the others. And everyone knows the logs of 2,3,5 and 7....right ???
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?