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Projectile Motion and velocity homework

  1. Sep 30, 2005 #1

    IB

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    Your original velocity is [tex] {9.6} {m/s^2}[/tex] but after it is increased by 5%, it's 10.08 [tex] {m/s^2} [/tex]. How much longer would the jump be?

    Should I use the range formula or what should I do?
     
  2. jcsd
  3. Sep 30, 2005 #2

    Diane_

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    Insufficient information, and some of it is contradictory. You say you're talking about velocities, but you're using the units of acceleration. You ask how much longer the "jump" would be - what jump? Are we talking horizontal distance, vertical distance, or time in the air? You'll need to be a lot more specific before anyone can help you.
     
  4. Sep 30, 2005 #3

    IB

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    An athlete executing a long jump leaves the ground at a 28.0 degrees and travels 7.80m. a) what was the takeoff speed? b) if this speed were increased by just 5.0%, how much longer would the jump be?

    I've got part a), which is 9.6m/s

    But for part b) i don't know what to do.
     
  5. Sep 30, 2005 #4

    Diane_

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    I'll assume you have part (a) right. For part (b), you know the takeoff speed: 1.05(9.6 m/s). You know the initial angle, and you mentioned having a formula for range - I'd say just plug those numbers into the formula.

    Do you understand where the formula comes from?
     
  6. Sep 30, 2005 #5

    IB

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    I'm not sure. Please show me.

    The range formula is: [tex] {R=} (v sin 2 (theta))/g)} [/tex]

    ETA: how to do the theta in latex?
     
    Last edited: Sep 30, 2005
  7. Sep 30, 2005 #6

    Diane_

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    I don't really want to derive the range formula here. That generally takes me two full blackboards and about fifteen minutes. Let me give you a precis:

    You launch something at an initial velocity of v0 at some angle Q. You can break that initial velocity down into horizontal and vertical components. If we neglect air resistance, the horizontal velocity will be constant - at least until the thing hits the ground. If we knew how long it was in the air, all we'd have to do is multiply the initial horizontal velocity by that time and we'd have it.

    The amount of time it spends in the air depends on gravity. We know its initial vertical velocity - all we have to do is figure out how long it'll take it to rise to the peak, stop, and fall back to Earth to have the time. There are two ways to do this: (1) we can figure the amount of time it will take it to slow from its initial upward velocity to zero under the acceleration of gravity. This will be the time to the peak. We then double it to get the total time in the air. (It will take just as long to fall back as it did to rise.) Or (2) We can use a relation between displacement, time, acceleration and initial velocity to solve directly. We have the initial velocity, the acceleration, and we know that if the object falls back to the ground, the vertical displacement will be zero. The relation we need will be quadratic in time, so we'll get two times. One will be zero - that's when the object was first at height zero, the start of the problem - and one will be the answer we need. Multiply that by the initial horizontal velocity and Bob's your uncle.

    Take the resulting formula and plug 1.05(9.6 m/s) = 10.08 m/s in for the initial velocity and 28.0 degrees for your angle and there you go.
     
  8. Sep 30, 2005 #7

    IB

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    So are we finding R? Is it: R = (10.08* sin(2*28))/(9.8)?
    For that I got R = .85m
    Is that how much longer if the velocity is increased by 5%?
     
  9. Sep 30, 2005 #8

    Diane_

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    Yes, you're finding R, but that doesn't look even approximately right Hang on - I'm going to have to work this one out myself. I'll post again in a couple of minutes.

    (Now I have to go derive the range equation. :) )
     
  10. Sep 30, 2005 #9

    IB

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    Why do you need to derive it? I've given you the range formula.
    :smile:
     
  11. Sep 30, 2005 #10

    Diane_

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    I should have remembered that. You lost a square in the range formula. It should be v^2, not just v.

    You should be sure you get these answers yourself, but I'm getting 9.60 m/s for the first (as you did, but watch your significant digits) and 8.60 m for the second.

    So, apart from losing the square in the second part of the problem, it looks like you're doing fine. Watch the significant digits, and keep track of your units. That will help you avoid problems like losing powers. :)
     
  12. Sep 30, 2005 #11

    Diane_

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    Side comment - I had to derive it because I wanted to be sure of it. Memory can play tricks on you, as mine and yours did. But if you know where the equations come from, you can always be sure of getting them right.

    And besides, it's good practice.
     
  13. Sep 30, 2005 #12
    OK - I have a projectile motion problem;

    A dart gun is fired while being held horizontally at a height of 1.00 m above ground level, and at rest relative to the ground. The dart from the gun travels a horizontal distance of 5.00 m. A child holds the same gun in a horizontal position while sliding down 45-degrees on an incline @ a constant speed or 2.00 m/s. How far will the dart travel if the child fires the gun when it is 1.00 m above the ground?

    I'm having trouble visualizing the diagram & solving....
     
  14. Sep 30, 2005 #13

    HallsofIvy

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    Sango89- It's a very bad idea to append a new question to the end of another thread. A lot of us note when a question has been answered and then don't look at that thread again. Much better to start your own thread!
    (Even if you were the person who started the thread, it's better to begin a new thread for a new question.)

    Okay, you know the initial height was 1 and the initial vertical speed was 0 ("being held horizontally). You know that the acceleration due to gravity is -9.8 m/s2. That tells you that the height h, at time t seconds after firing the dart, is
    [tex]h(t)= -\frac{9.8}{2}t^2+ 1.0[/tex]
    You can solve that for the time, t, when the dart hits the ground (h(t)= 0).

    You do not know the horizontal speed but, calling that "v", you how that the distance the dart travels, horizontally, is d(t)= vt. Since the dart hits the ground 5 m from the gun, d(t)= vt= 5 when t is the time you just found. Put that time in and solve for v.

    Finally! The child is sliding down 45 degrees at 2 m/s. Convert that to horizontal and vertical components and add to the initial velocity (v, 0) with v as you just found.

    Now you can use the same formulas as above, with these new initial velocity components to find the time when the dart hits the ground and the distance horizontally where it lands.

    Anyway:
     
    Last edited: Sep 30, 2005
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