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Questions about motion in 2-Dimensions

  1. Oct 9, 2006 #1
    Hey guys, have a few questions I was hoping you guys could help me out with.

    *

    4.13)******* One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown at an angle of 70.0 degrees with respect to the horizontal. (a) at what angle should the second snowball be thrown to arrive at the same point as the first? (b) How many seconds later should the second snowball be thrown after the first to arrive at the same time?

    What I did:

    First I drew a digram and made a line represting the resultant (r) vector which is 25.00 m/s. I knew my angle, which is 70 degrees. from this, I'm able to use one of 4 formulas to find the components of r which = 25 degrees.

    for velocity:
    Vxi = (vi)(Cos)(angle)
    Vyi = (vi)(Sin)(Angle) - gt

    for displacement:
    x = (vi)(Cos)(angle)(t)
    y = (vi)(Sin)(angle)(t) - 1/2(g)t^2

    so I solved the initial velocity for the x components, which was:
    vxi=(25)(Cos)(70) = 8.5 m/s

    and the y component

    vyi= (25)(sin)(70) = 23.49 m/s

    at this point, I tried a formula for the x component for displacement, and got:

    8.5t

    but I figured I could find time by using the y component seeing as how I know the snowball will be at 0 velocity at the max height. So I could use the kinematic equation vf = vi+at . so if I plug that in I get:

    0 = 23.46 + (-9.8)t

    -23.46/-9.8 = 2.3969 s

    and I would have to x2 so that it covers the whole distance so the t = 4.793 s

    now, if I plug that into the equation 8.5t , I would get :

    8.5 (4.793) = 40.7405m

    So now I know my displacement of snowball 1 (and I'm assuming for snowball 2)

    Now at this point I'm stuck. I don't know what to do with the information I have to get the missing angle for snowball 2 and have no idea how to calculate what I need to launch snowball 2 for it to = the time for snowball one. Can anyone help?



    My Next question:



    *4. 23) A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.



    What I did:

    Again I drew a diagram. I let my y=-40m from where my origin is. First I calculated the sound coming back to the soccer player, as that information was given in more detail. So I know the formula X=(Vi)(t) .
    So (343)(3.00) = 1024m. So I know my hypoteneuse and my opposite side which is -40. So if I apply the pith. theorm, I get:

    1024^2 - (-40)^2 = 1023.2184 m

    So now I know my displacement.

    At this point I want the total time from the kick to when it falls into the water so that I can calculate the initial velocity. I assume that I can use this formula:

    xf = vyi t -1/2 gt^2

    I assume that vyi is 0 because the soccer player kicks it straight and than it goes down (don't know if this is a valid assumption). Also, I subtract t-3 since we want to see when the rock hits the water, not includng the sound back to the the soccer player.

    -40 = 0 - 1/2 (9.8)(t-3)^2

    -40 = -4.9 (t^2-6t+9)

    -40 = -4.9t^2 = 29.4t -44.1

    = 4.9t^2 - 29.4t = 4.1

    since I have a quadratic equation, I use a quadratic formula which gives me a time of 5.8 seconds. which is essentially time of flight. So when when I find the intial velocity, i plug it into this equation:

    x = (v)(t)

    1024 = (v) (5.8)

    = 176 m/s


    ( I don't know, but I really doubt this answer for some reason, so if someone can correct it for me, it would be much appreciated!)





    Last Question:

    *

    4.54) A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 40.0 degree angle with the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the the backboard? The basket height is 3.05 m.


    Again I drew a diagram. I could subtract 3.05 from 2.00 to give me 1.05m. So now I figure I can use equations

    vxi = (vi)(cos)(angle)
    and
    vyi = (vi)(sin)(angle)

    I have as a result:

    10 = (vxi)(t)
    and
    1.05 = (vyi)(t) - 4.9 t^2

    As you can see, I'm missing a few things. but I go through the motions of multiply through what i can.

    for x: 10 = (vi)(cos)(40)

    for y: 1.05 = (vi)(sin)(40)t - 4.9t^2

    =

    10 =(vi)(0.76604)

    =

    10/0.76604 = 13.05 (Initional velocity)

    I than use the intital velocity and plug it into the y component equation. 1.05 = (13.05)(sin)(40)t - 4.9t^2

    1.05 = (8.38834)t - 4.9t^2

    =

    -4.9t^2+8.38t-1.05

    now that we have a quadratic equation, I put it through a quadratic formula to give me 1.5 seconds.

    so the equation I use for intitial velocity is x = v(t).

    10 = vxi (1.5) = 6.66 m/s

    and

    1.05 = vyi (1.5) - 4.9 (1.5)^2 = 8.05 m/s

    so once I have those two velocities, I use a^2+b^2=c^2 to find the hypoteuse which yields 10.4478 m/s for my initial velocity.



    Just to let you know, I am very new to physics, in fact this is teh first college physics course I am taking, so for all I know, all my answers could be wrong. If you guys can let em know the right way about going through problems, and tips or tricks, it would be much appreciated.

    If you guys can double check my answers and let me know, I would really appreciate that as well :)

    take care and thanks very much!

    Neeraj
     
  2. jcsd
  3. Oct 9, 2006 #2

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    It would be better if you posted separate problems in separate posts. I cannot make sense of the statement of the first problem, so I did not try to check your work. The sentence before part a) contradicts part a).

    Your second problem has at least one error. 3 seconds is how long it takes the rock to reach the water and for the sound to get back to you. It is not the time between the splash and when you hear it.

    I didn't look at any more.
     
  4. Oct 9, 2006 #3
    will do. I'll make 3 seperate posts. As for question 1, I double checked to make sure of any errors in my textbook. It's exactly as it was stated.
     
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