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Projectile motions

  1. Jun 20, 2004 #1
    I'm having trouble with these two problems. We haven't covered this in class yet, but I have to do them for my lab tomorrow.

    1. A package is dropped from a passing plane(moving in the horizontal direction) and produces a projectile motion as it is falling to the ground. If it takes the package only 50 seconds until it hits the ground, how far above the ground was the plane when the package was released? in meters. How do i know what equation to use for this?

    2. A ball rolls horizontally off the edge of a tabletop that is 1.5m high. It strikes the floor 2.80m away from the edge of the table. What is the initia (horizontal) velocity of the ball? in m/s. I don't know what equation to use for this either. There are so many in the text, but we haven't gone over problems like this yet. Please help!!
  2. jcsd
  3. Jun 20, 2004 #2


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    These are free-fall equations; can you set up Newton's 2.law in that case?
    Newton's 2.law of motion will remain (for a long time) the fundamental equation you, must try to solve, in one form or another.
    (Newton's 3.law helps you in many cases to identify what forces acts on various objects)

    Just about every other equation you'll meet for a long time be results gained from solving Newton's 2.law.
    If you don't remember a particularly useful result in a given situation, you must get back to basics (trying to solve Newon's 2.law to derive that result)

    So again, how would you set up Newton's 2.law of motion in the 2 cases?
  4. Jun 20, 2004 #3
    If something is moving horizontally when it falls, does it take longer to land?

    Give it a try - it doesn't. If you can neglect air resistance, all objects accelerate at g when dropped, no matter how fast they're moving perpendicular to gravity.

    So, for the first question, you can totally ignore the fact that the plane is moving. You know the object takes 50 seconds to land. Here are the only two equations you'll ever need for a constant acceleration problem (you can find the others by solving and substituting between these two)

    [tex]x - x_0 = v_0 t + \frac {1}{2} at^2[/tex]


    [tex]v = v_0 + at[/tex]

    Where [itex]v_0[/itex] is the initial velocity, [itex]x_0[/itex] is the initial displacement, t is the time, a is the acceleration, v is the current velocity, and x is the current displacement.

    Let's use the first one. Think of an x-axis sticking out of the plane and straight down into the ground, with the origin at the plane. Then, at the instant when the package is dropped, it has zero displacement and zero velocity (along the x-axis, we don't care about the plane's forward movement.) What about the acceleration? We normally approximate g (the acceleration due to gravity) as 9.8 m/s². In reality, it varies depending on your position on Earth, but 9.8 m/s² is close enough.

    So, what we know is:
    The package's initial displacement = [itex]x_0[/itex] = 0
    The package's initial velocity = [itex]v_0[/itex] = 0
    The acceleration due to gravity = a = 9.8 m/s²
    The time the package has been falling for = t = 50 seconds

    Putting this into our first equation gives:

    [tex]x - (0) = (0) \cdot 50 s + \frac {1}{2} \cdot \frac {9.8 m}{s^2} \cdot (50 s)^2[/tex]

    This simplifies down into x = 12250 meters. The package has been displaced by 12250 meters after 50 seconds. However, using such a specific number implies we know exactly how much it has fallen. We don't - we only know the amount it has fallen to the same precision as the numbers we used to calculate it in the first place. "50 seconds" probably has two significant digits, so let's round our answer off to two significant digits as well. 12250 meters is about 12000 meters, or 12 kilometers, or 1.2 x 10^4 meters.

    In the real world, it wouldn't be anywhere near that high, because the package would have long since hit its terminal air speed. Plus, planes don't generally leave the troposphere :wink:

    For the second question, you can find out how long the ball has been in motion by considering just the vertical velocity and acceleration, and then substitute that time to find the initial horizontal velocity. Give it a shot, and ask if you're still having problems with it.
  5. Jun 20, 2004 #4
    how do i find the vertical velocity? I guess I just don't understand what vertical and horizontal velocity mean? For question 2, you said to find the time by considering the vertical velocity and acceleration, but how do I do that? Thank you.
  6. Jun 20, 2004 #5


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    The easiest way to solve these problems is without having to resort to finding the right equation to plug numbers into. I'm not sure if you have studied projectile motion yet, but one thing to remember when doing this problem is that the horizontal component of the velocity is not changing throughout the flight.

    You can take advantage of this fact by realizing that the problem gave you the total horizontal distance traveled. Now if you can find the total time the flight took, you can just divide the distance by the time to find the velocity(since the average velocity IS the horizontal velocity through the flight, since this velocity isn't changing. The vertical velocity is changing however, due to gravity, but don't worry about it just yet). Hmmmm, I wonder where you can find that total time? Hint: Look at the height of the table and remember the acceleration due to gravity. The time the ball takes to reach its final vertical position is obviously the same time it takes to reach its final horizontal postion.
    Last edited: Jun 20, 2004
  7. Jun 20, 2004 #6
    is there an equation to find the time? we haven't studied this yet, and i really don't see how to find the time with just the height of the table and the acceleration. i can't find anything in the text to link those two things in order to find time.
  8. Jun 21, 2004 #7


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    No problem. I don't feel comfortable handing over an equation to you and telling you to plug in numbers, since that is a really bad way to learn physics. And i'm not sure of your familiarity with calculus, but just try to follow along and justify each step to yourself to help your understanding.

    We need a way to relate a vertical distance with a time. What we know is the acceleration due to gravity, and its relations to velocity and time:

    [tex]\vec{a} = \frac {dv} {dt}[/tex]

    We can forget about vectors, since we are only talking about the motion in one dimension (just vertical), and "multiply through" by dt to obtain:

    [tex]a dt = dv[/tex]

    now integrate it:

    [tex]\int_{0}^{t_f}a dt = \int_{v_0}^{v_f}dv[/tex]

    this gives us:

    [tex]at_f = v_f - v_0} [/tex]

    we know that v0 is zero, since we want the the total time of the fall, and this time really starts when the velocity is zero. (we are completely ignoring anything happening in the horizontal direction) call t_f, just t, call v_f just v, and lets remember the definition of velocity, and integrate one more time to get what we want.

    [tex]\vec{v} = \frac {dx} {dt}[/tex]

    plugging back into the last equation

    [tex]at = \frac {dx} {dt} [/tex]

    multiplying through one more time by dt

    [tex]at dt = dx[/tex]

    now integrate again...

    [tex]\int_{0}^{t_f}at dt = \int_{x_0}^{x_f}dx[/tex]

    to get...

    [tex]at^2 = x [/tex]

    this is the relation your problem calls for, between the distance the ball travels vertically and the time it takes. Notice I made x0 simply zero to make life easy. To avoid this kind of derivation every time you solve a simple kinematics problem, I would recommend simply memorizing the kinematics equations you are studying, and knowing the ins and outs of how to derive and use them. You're welcome :smile:
  9. Jun 21, 2004 #8
    Just out of curiosity, akatsafa, how do you have a problem on Newton's Law of Gravitation and forces before you've studied kinematics?

  10. Jun 21, 2004 #9
    the lab and the lecture do not coincide....so the lab is always ahead of the lecture.
  11. Jun 21, 2004 #10
    Oh. Bummer.

  12. Jun 23, 2004 #11
    1) Resolve vertically (take the vertical components at hand, horizontal components can be disgarded as they will not affect downward motion, only the horizontal displacement).

    U=0 (initial downward speed is 0 as the plane is moving horizontally).
    a=9.8 (gravity)
    t=50 (seconds)
    S=? (this is your displacement in metres)

    S = ut+.5at^2
    = 0x50 + .5x9.8x50^2
    = 12250m

    Resolving vertically:
    S = 1.5, u = 0, a=9.8, t= ?
    1.5 = .5x9.8xt^2
    1.5 = 4.9t^2
    t = 0.553283seconds. This is the time to reach the floor

    Now considering the vertical components:
    u = ?, a = 0, t =0.553283seconds, S = 2.8
    s = ut + .5at^2
    2.8 = 0.553283u
    u = 5.061m/s

    Although i don't study physics i study maths, here are some of the kinematics equations you'll often find yourself using:
    These equations are only useful (as far as i am aware), when acceleration is a constant, this occurs in a large number of situations. When acceleration is not constant you'll need to know one of these as a function of time, and consider a speed time graph in differentiation from displacement to velocity and velocity to acceleration.
    S = u.t + 1/2.a.t^2
    V = u+at
    V^2 = U^2 + 2as
    S = .5(u+v)t

    Where S = Distance.
    U = Initial speed.
    V = Final speed.
    a = Acceleration
    T = Time.

    I would also advise you to operate in m, seconds and m/s.
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