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Projectile motion help

  1. Oct 30, 2009 #1
    1. A coin is flicked horizontally off the top of a table. The diagram shows the horizontal and vertical components of the velocity after 0.45 s. What is the speed of the coin at this instant, given by the red vector?



    2. Okay im rerally not sure how to do this. There is a diagram of two lines one going in teh horizontal direction and one in teh vertical so tehy have made a right angle and then a red line going diagonaly down through the right angle. I don't know what eqaution to use but I was thinking of when you use cos or sin to find teh horizontal or vertical components but I don't know where gravity and time fit into that?!



    3. Im not sure how to do it.
     
  2. jcsd
  3. Oct 30, 2009 #2
    You can find any resultant velocity by forming a right-angled triangle with the verticle and horizontal components of velocity and using Pythagoras' Theorem.

    You know that the horizontal motion of the coin is denoted by the equation distance = speed * time, and you're given the time in the question, and the speed (the horizontal component of the velocity). You can use this to work out the distance from the table the coin is after 0.45s.

    Finding the verticle velocity is slightly more tricky as you need to take into account the acceleration due to gravity. Remember that since the coin had only a horizontal velocity as it was flicked off the table, the initial verticle velocity is 0.

    Once you've got both velocities, form a triangle as I said above and you should then have the overall velocity. If you're asked for the angle as well, you can use trigonometry on the triangle.
     
  4. Oct 30, 2009 #3
    Ahh okay so I worked out the distance teh coin went which is 3.15?

    Waht I forgot to say was that there was a 7 ms^-1 on teh graph by the horizontal component. So I worked out the distance the coin went with that and got 3.15 should I have used 9.81? If I use 9.81 then I get 4.41 :)

    What equation do I use for the vertical velocity? And there is no angle :)
     
  5. Oct 30, 2009 #4
    Since the initial horizontal velocity is 7m/s then yes the distance after 0.45s is 3.15m.

    Remember that horizontal motion is not affected by gravity, so there is no need to factor this into your equations. All you need is distance = speed * time.

    For verticle velocity, you can use a number of equations but the one most appropriate here is:

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

    Where v = final velocity, a = acceleration (in this case 9.81ms^-2), t = time and u = initial velocity.

    In this case remember that initial velocity in the verticle direction is 0.
     
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