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Homework Help: Projectile Motion question. Help.

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown from the top of a 20 m high cliff. It strikes the ground at a certain angle. Consider the base of the cliff to be at height 0 m. The horizontal distance is measured along this level. If the ball is thrown horizontally, and it strikes the ground at 45 degrees, what was it's initial speed?

    2. Relevant equations

    vfy=vosintheta -gt

    3. The attempt at a solution
    I thought that if it strikes the ground at 45 degrees, the initial horizontal and vertical components of velocity must be the same. But then someone told me that this meant that the final vertical and horizontal components were the same. So I used gt and pythagorean theorem and got the answer, but I'm not sure why the final components are equivalent if the vertical component has gravity included and the horizontal does not. Please explain. Also, I'm not sure how finding the final components helped solve for the Initial velocity. Please Help.
  2. jcsd
  3. Jan 28, 2009 #2


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    The problem states that the ball is thrown horizontally. If this is true, how can the initial horizontal and vertical components of velocity be the same?
  4. Jan 28, 2009 #3


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    Welcome to PF.

    Yes. It will strike at 45° when the component of horizontal = component of vertical - 45°.

    That means then just figure the vertical velocity it will hit the ground if it was dropped straight down. Then if it's launched at that speed ...
  5. Jan 28, 2009 #4
    well, how can the final components be the same if the vertical has gravity involved and the horizontal doesn't?

    and if i were to figure the vertical velocity, it would just be -gt, which i have calculated to be -19.6 because t=2. then to find the initial velocity would it just be sqrt of 19.6^2+ 19.6^2? How is this the initial velocity when we are using the final components? Sorry I am asking so many questions, i just really want to understand this.
  6. Jan 29, 2009 #5


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    You need to pay careful attention to the wording.

    At launch there is only horizontal velocity. It won't change through out the fall.

    The problem then tells you it hits at 45°

    That tells you that the vertical velocity and the horizontal velocity at the bottom are the same.

    But what is the vertical velocity after falling that height. Vertical only is all we want because we are working with the components independently.

    V2 = 2*9.8*20m = 19.8 m/s

    Since you know the horizontal didn't change from launch AND it must be the same as the vertical, then you know your INITIAL velocity.
  7. Jan 29, 2009 #6
    I get everything except for the very last part..how come you didn't use pythagorean theorem at the end to figure out the magnitude of final velocity?
  8. Jan 29, 2009 #7


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    But don't be so surprised that you got it.

    It's not rocket science.

    Oh wait. Maybe it is.

    Good luck anyway.
  9. Jan 29, 2009 #8


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    Btw, if they wanted to know FINAL velocity then you would use the Pythagorean relationship to find the |velocity| at impact.

    But that's not necessary from the statement. The 45° constraint - which is what they give you, not the velocity of impact - already gives you all the information you need, because it tells you the horizontal component must equal the vertical at impact.
  10. Jan 29, 2009 #9
    Explain again why you can use the vf2 equation because that deals only with the vertical....since delta y =20. Why cant I just say, the initial component of velocity in the x direction = -19.6, the initial component of velocity in the y direction = 0. So then sqrt of 19.6^2 + 0^2 = 19.6
  11. Jan 29, 2009 #10
    and sorry, but once again. could you explain how you know at 45 degrees the final horizontal and vertical components (I know sin and cos are the same), but how can you equate when the vertical includes gravity
  12. Jan 29, 2009 #11


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    Well that equation works either way so long as the Vf2 - Vi2 = 2*a*Y

    The presumption is that the difference in the squares of the velocities is twice the distance times the acceleration. It is not a hard equation to derive and one worth doing. (Remember: The square root of a negative is imaginary so if one side is negative then the final and initial are undoubtedly reversed, because if a is positive the final must be greater than the initial.)

    As to why 45° as opposed to the |V| at impact ... well that's the statement of the problem. If it was |V| that was = Vx, then you would have a different calculation. In fact it would be one that can't be made because it is impossible. As soon as the ball begins to drop it is already going faster than Vx for exactly the reason you understand.

    If they had wanted to ask the same question, but not give the angle of 45° they could have also asked how fast it was initially going if it struck the ground with a SPEED of Vf = √ 2*Vi then that would yield the same result.

    But they didn't. If they had given a different angle you could simple relate tanθ = Vy/Vx.

    In our case tan45° = 1 which gave us the Vx = Vy directly.
  13. Jan 29, 2009 #12


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    Whether one component includes acceleration and the other not, doesn't really matter. It's what is it at the instant it hits. You are only equating instants in time, not equations. This was why we had to calculate what its V in the Y direction was when it hit.
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