Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile in Motion: horizontal

  1. Aug 26, 2004 #1
    i'm having trouble with this problem, so i want to share it with you guys so some of you can briefly describe what is it asking for and how to solve using what fomula.... the question is

    A ball is fired from the ground with an initial speed of 1.70 x 10^3 m/s (which is approximately five times the speed of sound) at an initial angle of 55.0 degree to the horizontal. Neglecting air resistance, find the following

    A. the ball's horizonal range
    B. the amount of time the ball is in motion...

    initial Velocity = 1.70 x 10^3
    initial angle = 55.0 degree

    what formula should i use?
  2. jcsd
  3. Aug 26, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is it asking for?

    Part A is asking how far the projectile moves horizontally from its initial position

    Part B is self explanatory.

    Think: what do you know? You know that, by definition, a projectile is a flying object that, once given an initial velocity, undergoes motion governed solely by the force of gravity. Hence, this is a constant acceleration problem. Which kinematics formulas should you apply to such a problem?
  4. Aug 26, 2004 #3
    There are two velocity components : horizontal velocity ( 1.70 x 10^3 x cos55 ) that remais the same, and vertical velocity (1.70 x 10^3 x sin55 ) that is modified by the gravity acceleration along the flight.
  5. Aug 27, 2004 #4
    here is the derrivation of the formulas, you can plug in the numbers yourself:
    [tex] d_{V} = Vsin \theta t + \frac{1}{2} at^2 [/tex]

    [tex] 0 = (Vsin \theta) t - \frac{1}{2} gt^2 [/tex]

    [tex] 0 = Vsin \theta - \frac{1}{2} gt [/tex]

    [tex] t = \frac{2Vsin \theta}{g} [/tex]

    this is the formula for total flight time, now you must use this formula to sove for distance:

    [tex] d_{H} = Vcos {\theta}t [/tex]

    [tex] d_{H} = Vcos {\theta} (\frac{2Vsin \theta}{g}) [/tex]

    [tex] d_{H} = \frac{V^2 sin2{\theta}}{g} [/tex]

    now just sub in your numbers and you should be ok.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook