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Projectile Motion Problems

  1. Sep 26, 2012 #1
    1. A soccer ball is kicked so that it has a range of 30m and reaches a maximum height of 12m. What velocity (magnitude and direction) did the ball have as it left the footballer's foot?

    I used the equation to solve for time and got 1.55s. I'm not sure if air resistance is involved, we never discussed how to solve problems with air resistance in class.

    2. A stone is thrown with a speed of 20.0 m/s at an angle of 48° to the horizontal from the edge of a cliff 60.0m above the surface of the sea.
    (a) Calculate the velocity with which the stone hits the sea.
    (b) Discuss qualitatively the effect of air resistance on your answers to (a).




    3. The maximum height reached by a projectile is 20m. The direction of the velocity 1.0s after launch is 20°; find the speed of launch.

    Is air resistance included in all of these problems?
     
  2. jcsd
  3. Sep 26, 2012 #2

    lewando

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    For the first problem, what does 1.55s represent? Air resistance is assumed negligible.
     
  4. Sep 26, 2012 #3
    Sorry, s stands for seconds. Could you explain how to solve the problems?
     
  5. Sep 26, 2012 #4

    lewando

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    Sorry, I meant the interval of time, in seconds, what does that represent? Which equation did you use? Show those steps. For example of what I'm asking: does your time result represent the total flight time or the time it took to reach the highest point during flight?
     
  6. Sep 26, 2012 #5
    I think the time is how long it was in flight, I'm not sure if this is correct or if I was even supposed to solve for time. I used this equation:

    Δy=Vyt + 0.5αyt^2
    12=0.5(-10)(t^2) and used this to solve for t
     
  7. Sep 26, 2012 #6

    lewando

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    Okay, there was a term, Vyt. How did you make that disappear?
     
  8. Sep 26, 2012 #7
    Isn't the velocity of y zero? So then it would be (0)(1.55) which is zero, right?
     
  9. Sep 26, 2012 #8

    lewando

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    So to visualize what is going on-- along the y axis-- initially the footballer's foot gives an initial Vy0 resulting in the ball going up. At some time, the ball stops going up-- Vy = 0. This is what made Vyt disappear in your equation. You have solved for this time. Now the ball comes back down. The moment before it hits the ground, it has a non-zero velocity. Do not consider this zero velocity. By symmetry you can determine the total flight time.
     
  10. Sep 26, 2012 #9
    Ok then how do I use this to find the magnitude and direction of velocity?
     
  11. Sep 26, 2012 #10

    lewando

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    Now that you know the total flight time, can you determine Vx?
     
    Last edited: Sep 26, 2012
  12. Sep 26, 2012 #11
    I could use Vx=Vox+at but I don't know what Vox is.
     
  13. Sep 26, 2012 #12

    lewando

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    In the x direction, acceleration is 0. How about distance = rate * time?
     
  14. Sep 26, 2012 #13
    I'm not sure what that formula is, will it help me solve for velocity? These are IB questions, not sure if that makes a difference.
     
  15. Sep 26, 2012 #14

    lewando

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    That formula is a result of this formula with a = 0:

    x - x0 = v0t+ 0.5at2
     
  16. Sep 26, 2012 #15
    I think should start with what you know and do not know that lead to the answer.

    Knowns(3) - x and y(0 and 12m) distances.
    Unknowns(3) - v0, θ, t

    So you need 3 equations.
    There are 3 SUVAT equations.

    s=v0t + 1/2 at2
    v2=u2+2as
    v-u =at
    In this problem we are taking up positive and acceleration a as negative.

    You have 2 s(x and y) which are independent.
    So you need 3rd equation. You can use the last equation(vy-uy)=at as the third equation(only in y direction with acceleration).
     
    Last edited: Sep 26, 2012
  17. Sep 26, 2012 #16

    lewando

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    I need to check out for the night. I was going to suggest that you find Vx0 (or Vx, since ax is 0), and then Vy0. The vector sum will give you the answer.
     
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