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Kinematics questions. (Linear & projectile)

  1. Feb 25, 2013 #1
    Pictures of the questions are attached below.

    For question 2, I totally cannot get any answer from the four choices after I substituted everything in the equation.

    For question 4, I used V^2= U^2 + 2AS, with U being zero, A being 9.8 and S being 2.0. I got C but the answer is D. Can someone explain it to me?

    For question 5:
    Firstly, why can't I calculate theta by directly using tan θ = 30/30 which makes θ=45°?
    And how do I get the answer?

    I have used (u sin θ)/9.8 = 1.2 but there are still two unknowns (u and θ). I can't find another equation to do simultaneous.

    Big thanks to anyone who can help me solve these.

    p/s: those are A-levels questions. :blushing:

    Attached Files:

  2. jcsd
  3. Feb 25, 2013 #2

    Doc Al

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    What equations did you use? (Note that you can immediately eliminate several of the choices based on units.)

    The initial velocity isn't zero.

    The angle formed by that 30/30 right triangle is not the same as the angle of the trajectory. (It would be if the ball just went straight, but it doesn't.)
    That's one of the equations you'll need. Find another relating to distance.
  4. Feb 25, 2013 #3
    s= ut+ 1/2 at^2.

    Based on units, the choices left are A and B.

    So is that equation correct?


    The helicopter goes up with a steady velocity, so it's motion is controlled, not free fall or projectile motion right? So how can I determine it's velocity when it's at a height of 2.0m?

    Furthermore, when the commando jumps off it, why isn't his velocity zero? How come the commando have a velocity the moment he jumps off?

    I used the range equation, and I found that u^2 sin 2θ = 588.

    Then I made it: 2 u^2 sin θ cosθ = 588, then divide it by u sin θ = 11.76

    so u cos θ = 25, which makes u tan θ = 25/11.76.

    ∴ θ= 25.2. Is this the way?

    Last edited by a moderator: Feb 25, 2013
  5. Feb 25, 2013 #4

    Doc Al

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    Yes. Hint: Apply it twice.

    You are given the velocity of the helicopter.

    Because he is moving along with the helicopter.

    That's a fine way to do it.

    You can also use an equation for the horizontal distance:
    u cosθ 1.2 = 30.

    That's because you messed up the quotes.
  6. Feb 25, 2013 #5
    New questions

    For the objective question, the answer is D.
    I understand why it is negative, but I don't know why it needs a little time for the acceleration to become zero. Notice that the dotted lines are not parallel. What happened to the motion of the ball the time when it's going to touch the ground?

    For the structured question, it says velocity decreased from 30m/s to 22m/s, and distance is 26m. So I used the area under graph: 1/2 x (30+22) x t = 26, which makes t=1. My understanding is that the deceleration took 1s. Then, I calculate the remaining displacement by using the remaining area under graph: 1/2 x 4 x 22 = 44m. How come the answer is 30m?

    Attached Files:

  7. Feb 25, 2013 #6

    Doc Al

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    Re: New questions

    When the ball touches the ground, the ground exerts an upward force on the ball to change its velocity. The force isn't infinite, so it takes some short time for the ball to rebound off the floor.

    Looks good. So what's the acceleration?
    Where did you get the 4 from? Figure out the time it takes to come to rest.
  8. Feb 25, 2013 #7
    Re: New questions

    When the ball touches the ground, the ground exerts an upward force on the ball to change its velocity. The force isn't infinite, so it takes some short time for the ball to rebound off the floor.

    So what's with the slightly slanting part? Is it because of the friction between the ball and the ground that makes its acceleration decrease? (This idea is stupid, but)
    At the exact moment when the ball touches the ground, isn't the acceleration going to become zero straightaway? And remain zero throughtout the short impact time?

    Looks good. So what's the acceleration?

    Ah, I got it. By using formula. But can you explain why it isn't reasonable to directly calculate the area under graph? It's the distance travelled anyway, no?

    Where did you get the 4 from? Figure out the time it takes to come to rest.

    Rest time 30 - time after deceleration 26.

    Because he is moving along with the helicopter.

    But, when he jumps off, isn't the only force acting on him G?(ignoring air resistance)
    I mean, when you jump out of the plane at a height of 2.0m as stated, you should jump out exactly at a height of 2.0m right? If you consider the helicopter continues going up during his jumping time, then the height won't be 2.0m anymore. And even so, how does it affect his initial velocity? The helicopter goes vertically up and the person jumps vertically down, how come his velocity is not completely independent of the helicopter? Does he jump from a door below the helicopter or from the edge of the helicopter? And do they make any differences? (sorry :blushing:)
  9. Feb 25, 2013 #8

    Doc Al

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    Please make proper use of quotes. (Using color to distinguish quoted text is not a good idea.)
    Last edited: Feb 25, 2013
  10. Feb 25, 2013 #9

    Doc Al

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    Re: New questions

    No. The acceleration of the ball is not zero during the collision (except for when the ground force exactly balances the weight of the ball, for perhaps an instant). Realize that the ball is squishy. It gets compressed as it rebounds. As soon as it loses contact with the ground, once again gravity is the only force acting.

    Nothing wrong with calculating the area under the graph, if you do it correctly.

    Take another look. It comes to rest before time = 30.

    After he jumps, only gravity acts.
    Assume he jumps when the height is exactly 2.0 m.

    If step out of a car moving at 60 mph is your speed independent of the car? At the moment he jumps, his speed is exactly that of the helicopter. (Assuming he doesn't shove off and give himself extra speed.)

    No difference.
  11. Feb 26, 2013 #10
    Thank you very much. Really helps me a lot.
    Can I just post in this thread everytime I have questions?

    1. If ball X is dropped vertically off a height and ball Y is thrown horizontally from the same height at the same time, why will they reach the ground at the same time? Isn't Y's path longer?

    The following two questions are attached together with my workings.

    2. Value of T2 is totally unreasonable. How should I arrange the forces? Since the two angles are both 45°, I can either put tan 45 = T2/100 or tan 45= 100/T2. How could that be?

    3. Is my working correct? I don't know how to continue. I cannot just substitute e=1/2 into the equation, right? Even if I do, I still couldn't get the value of Va.

    (here is the 4rd picture)

    Attached Files:

  12. Feb 26, 2013 #11

    Doc Al

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    You are welcome.

    You'll probably get help a lot quicker if you post a new thread. We have many top notch helpers here.

    The path may be longer but it's also moving faster. The only thing that counts towards the time it takes to reach the ground is the vertical motion. And since both have zero initial speed in the vertical direction, they will both fall at the same rate.

    Any horizontal speed only makes it go sideways as it falls. Who cares?

    Well, what does tan 45° equal? Does it matter?

    How did you get a value of T2 = 0.01 N? Check your work.

    Unfortunately, I cannot view the picture you linked.
  13. Feb 26, 2013 #12
    All solved. Bless you.

    Here it is.

    Another question: (Since it's still projectile motion, I might as well continue here)

    Please view the second image.

    I have problems with d) and f).

    For d), I have found the t above the line to be 1.53s. For the part below it, I tried two ways:

    1. Vertical component of velocity: 15sin30
    t = 15sin30/G = 0.76

    2. Height: 50m
    t= 50/(15sin30) = 6.67 (?!)

    Please explain to me why it's wrong.

    For f), I calculate the distance by adding the range of the projectile motion and the horizontal component of the remaining trajectory.

    Range = (U^2 sin 2θ)/g = 18^2 sin 2(30)/9.81 = 28.6m
    Horizontal component = vt = 18cos30(2.52) = 39.3m

    It's wrong. The real answer is 72.9.

    *The answer for d) is 4.05s. I minus 1.53 from it and got the 2.52.

    Thanks in advance.

    Attached Files:

    Last edited: Feb 26, 2013
  14. Feb 26, 2013 #13

    Doc Al

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    Your work is fine. Express Va in terms of u. What minimum value of e will give Va a negative direction?
    You can express Va and Vb in terms of u.
  15. Feb 26, 2013 #14

    Doc Al

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    Method 1 would find the time for something to reach that speed start from zero. (That's the time it takes to rise to the top.)

    Method 2 assumes a constant speed for the descent.

    Neither method will work. Why not just use:
    [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]

    The horizontal speed is not correct.

    And the easy way, once you've found the total time, is just distance = vt (for the horizontal motion). No need for the range equation.
  16. Feb 27, 2013 #15
    Now I use v^2=u^2 + 2as, where v is the vertical component, u = 15sin30, a=9.8, s=50.
    Then I substitute the V value into a= (v-u)/t. t= 2.52s.

    Is this the way, or are there any better ones?

    What does the o mean?
    Is this derived from the s=ut + 1/2at^2 formula?

    Ooh, it's 18 m/s.
    But (18 x 2.52) + 28.6 is still not the answer. Which part is wrong?

    And, please see the image again. I thought I would easily solve e) after I finished d) and f) but :uhh:

    I used v^2 = u^2 + 2as for vertical component.
    v^2 = (15sin30)^2 + 2(9.8)(50), v= 32.2 m/s

    Then to find the actual velocity, I calculate v' cos (90-30)= 32.2, which makes v' 64.4m/s which is far from the answer :frown:
  17. Feb 27, 2013 #16

    Doc Al

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    That's one way. I would use the formula that I provided.

    The 0 signifies initial position and initial velocity. (As in T = 0.)
    Yes, same thing.

    I believe you misapplied the range equation.

    But once you have the total time, finding the horizontal distance is easy without needing the range equation.

    That's fine.

    You found the vertical component of the final velocity. What's the horizontal component?
  18. Feb 28, 2013 #17
    R= (u^2 sin 2θ)/g

    = 15^2 sin 2(30)/9.81 where is wrong?

    And why is the time taken to attain max height (15sin30)/g, regardless of the real height? Wouldn't that be assuming that the ball just went straight?

    The other questions are solved. Tq.

    *And why is the vertical component of a point at the other side of the trajectory = 0? By other side I mean the part on the right of the symmetrical line.
    Last edited: Feb 28, 2013
  19. Feb 28, 2013 #18

    Doc Al

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    Wrong speed and wrong angle. (You need the angle with respect to the ground.)

    The height reached depends only on the vertical motion, which is why you use the initial velocity in the vertical direction.

    I don't understand your question. Are you asking about position (height) or velocity?

    Give me a specific example of what you mean.
  20. Feb 28, 2013 #19

    Its a projectile motion diagram, an upside down U letter.

    There's a point X on the left half trajectory. It's horizontal component is U cos θ, and vertical component < u sin θ. And there's a point Y on the right half of the trajectory(the path the object falls down), it's horizontal component is still ucosθ, but it's vertical component is 0.
  21. Feb 28, 2013 #20

    Doc Al

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    How can the vertical component be zero? The only place where the vertical component of the velocity is zero is at the apex of the trajectory.
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