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Vertical Projectile Motion, Please Check My Answers

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    3.1) 5m.s^-1.........Not sure about that

    3.2) vf^2 = vi^2+2*a*deltax
    = 1.5^2 + 2(-9.8)*300
    = 76.67 m.s^-1 down

    3.3) Wnet = Ek
    (0.392*300) + Wf = 0.5*0.04*(76.7-60)^2
    Wf = -112J

    3.4) Without air friction:
    (U+k)top = (U+K)bottom
    0.04*9.8*300 = 0.5*0.04*76.7^2
    117.6 = 117.6

    Mechanical energy is conserved when there is no external force such as friction acting on the system.

    Can someone please check these answers?
    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 2, 2013 #2


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    I think 3.1) is -5. The others look fine.
  4. Oct 2, 2013 #3


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    That looks right to me. :approve: Just be sure to specify the direction.

    For your initial velocity, are you sure you want to use the velocity of the marble relative to the parachutist? Since you are asked to find the velocity at which the marble strikes the ground, wouldn't it make more sense to use the initial velocity with respect to the ground as your initial velocity?

    I'm not following you there. I'm not even sure where the 0.392 figure comes from.

    Also, you're missing some square operations on the right hand side, one way or another. KE = (1/2)mv^2.

    Qualitatively speaking, that's correct.

    You're numbers on the other hand aren't quite right. The marble does have a little kinetic energy at the time it was thrown (at the top) in addition to the potential energy.
  5. Oct 2, 2013 #4
    Yes, but shouldn't you use the actual velocity that it was thrown up at?, the velocity relative to the ground is just the perceived velocity seen by that person on the ground, right?, eg, if a car moves at 30 m.s^-1 and another car moves at the same speed parallel to the other car their relative velocities are 0 but they are still moving at 30 m.s^-1.

    That would be the gravitational force Fg = mg = 0.04(mass of marble)*9.8
  6. Oct 2, 2013 #5


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    Velocity is always relative. Technically, both figures are correct; neither is the "actual" velocity or just the "perceived" velocity. But what you must do is specify what is relative to what, and then maintain consistency.

    I suggested using the velocities relative to the ground because it ends up being easier that way, and secondly, the 60.0 m/s in part 3.3 is pretty clearly the velocity relative to the ground, not relative to the parachutist. If you were to keep everything relative to the parachutist, you would also need to convert this 60 m/s velocity to be relative to the parachutist. But I'm pretty sure that seconds 3.2 and 3.3 are asking for the velocity and work relative to the ground. So it makes more sense to me to keep things relative to the ground for this problem.

    Okay, but I'm not sure how that fits in. All you really need is the difference in final, kinetic energies when friction is ignored and when it is not. :wink: (In both cases, whether friction is ignored or whether it is not, they have the same initial potential energy and initial kinetic energy. The final kinetic energies are what differ.)
  7. Oct 2, 2013 #6
    Okay, so I should use 5m/s.

    3.2) vf^2 = 5^2 + 2(-9.8)*300
    vf = 76.52m/s

    For 3.3)

    Kinetic energy without friction:

    1/2*0.04*(76.52^2 - 5^2)
    = 116.61 J

    Kinetic energy with friction:

    1/2*0.04*(60^2 - 5^2)
    = 71.5 J

    Work done on friction would be 116.61-71.5 = 45.11 J , Is that correct?
  8. Oct 2, 2013 #7


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    But is it 5 m/s up or down?

    Is the marble's initial velocity in the same direction as the gravitational acceleration or the opposite direction?

    There's no need to bring the 5^2 back into the picture. An object's kinetic energy is a function of its present speed (which you've already calculated or were given) not its past speed. The speeds are already relative to the ground and there's really no need to convert the kinetic energies to be relative to something moving at 5 m/s relative to the ground.

    That said, one additional thing you do need to change is the 76.52 number (see above).

    You have the correct idea here, but you'll need to change some numbers in previous steps so that this step comes out right.
  9. Oct 3, 2013 #8
    5m/s up. I am taking all upward motion as positive.

    Opposite. The ball is thrown upwards and gravity acts downwards.

    That vf should be 76.52m/s down.

    With or without the 5^2, I still get the same answer when I subtract them.

    Is 76.52m/s wrong?, or are you referring to the direction?, I know it is 76.52m/s down or -76.52m/s but it doesnt make a difference when you substitute it because you are squaring it
  10. Oct 3, 2013 #9


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    You might want to think about that.

    The parachutist's velocity is -6.5 m/s relative to the ground. The parachutists throws up a marble at 1.5 m/s relative to the parachutist. So what's the velocity of the marble relative to the ground? Is it positive or negative?

    If you're falling down, really, really fast relative to the ground, and you bump an object up relative to yourself, every so slightly, so it's not falling down quite as fast as you, but almost; is it still falling down or is it rising up, relative to the ground?

    So if you are falling toward the ground at 6.5 m/s, and you throw an object up at merely 1.5 m/s relative to yourself, is the object heading toward the ground or away from it?

    That's not quite the answer that I came up with.

    Yes, that's right! :approve: the solution works out whether you calculate the kinetic energies relative to the ground or relative to the parachutist! [Edit: Or any other relative velocity, such as one 5 m/s relative to the ground.] I won't argue with that. :smile: [But you must treat all relative velocities consistently.]

    My only contention is the previous part that I discussed above.

    Well 76.52m/s is not the final velocity of the marble relative to the ground, ignoring air resistance. That's my point.
    Last edited: Oct 3, 2013
  11. Oct 3, 2013 #10
    So it should be 5m/s down

    I dont see why I am not getting your answer, I am using this formula, vf^2 = vi^2+2*a*deltax.
    Vi = -5m/s and a = -9.8
  12. Oct 3, 2013 #11


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    Since a is in the same direction as your initial velocity, make it positive. :wink:
  13. Oct 3, 2013 #12


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    Or if you wish, keep a negative, meaning down is negative, but you must also realize that deltax is also negative, since the distance is down.

    This particular kinematics equation, vf2 - vi2 = 2aΔx requires a bit of additional thought compared to the other kinematics equations, because the square of the negative number becomes a positive number. So you need to ensure that things make sense when you set things up.

    [Edit: and for this particular problem, realize that if you keep velocities and distances relative to the ground, then the initial velocity, the acceleration, the final velocity and the change in displacement, all point in the same direction.]
  14. Oct 3, 2013 #13
    Ok, I am going to keep down as negative. My mistake, should have realised delta x is negative.
    vf^2 = -5^2+2*-9.8*-300
    = 76.84 m/s

    For 3.3) 118.1-72 = 46.1 J

    Using the work energy theorem I also found that I got the same answer
    Wg - Wf = 1/2*m*v^2..............Wg is the work done by gravity
    117.6 - Wf = 0.5*0.04*(60^2 - 5^2)
    Wf = 46.1 J

    Is this problem finally solved?, although this relative velocity is still bugging me abit. You mean like if a person where standing on the ground it would seem that the ball was thrown down when it was actually thrown up?. A bit confused
  15. Oct 3, 2013 #14


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    That looks right to me. :approve:

    But just to illustrate the limitations of this particular kinematics equation, you have,

    [tex] v_f = \pm \sqrt{v_i^2 + 2a \Delta x} [/tex]
    As you can see, you can't tell just by looking the equation whether the final velocity is positive or negative. But knowing that negative is down (you chose to define it that way), and knowing that the final velocity wouldn't make any sense if it were going up (for this particular problem), you choose the negative value, making the final velocity -76.84 m/s.

    That also looks correct to me. :approve:

    Again, it's all relative.

    If a car is heading North, and the passenger in the car tosses a sandwich out the window, the passenger would notice that the sandwich tumbles on the road behind him, South of him. But would you, standing on the sidewalk, see the sandwich continue tumbling North, albeit at a lower speed than the car, or would you see the sandwich suddenly shoot South of you?
  16. Oct 4, 2013 #15
    It would continue travelling North, maybe?
  17. Oct 4, 2013 #16


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    Yes. :smile: So the sandwich is tumbling North relative to you and the ground. But the passenger in the car observes it moving South relative to him.

    In this problem, the parachutist descending toward the ground at 6.5 m/s throws the marble up at 1.5 m/s relative to himself. So relative to the ground the marble is traveling at 5.0 m/s, toward the ground.

    Notice that if you were observing this from the ground you would see the marble is still falling toward the ground. The initial speed of the marble isn't as fast as the parachutists speed -- you observe the parachutist descending toward the ground 1.5 m/s faster than the initial speed of the marble -- but everything is still moving in the direction toward the ground.
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