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Projectile motion ball hit

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    These are the exact words:

    A player is ready to throw a ball with max. possible velocity of 15 m/s from a height of 1.5 m above the ground. He wants to hit the wall 16 m away from him at its highest point. If the ceiling height of the room is 8 m, find the angle of projection of the ball and the height of the point where the ball hits the wall.

    2. Relevant equations


    3. The attempt at a solution

    The attached file is the diagram.

    To find α, I did this:



    ## \therefore 15\sin(\alpha)=\frac{9.81\times16}{15\cos(\alpha)} ##

    ## \therefore 15^2\sin(\alpha)\cos(\alpha)=9.8\times16 ##

    ## \therefore \sin(2\alpha)=\frac{2\times9.81\times16}{15^2} ##

    ## \therefore \sin(2\alpha)=1.3952 > 1 ## <-- WHAT

    Did I do something wrong?

    Attached Files:

  2. jcsd
  3. Oct 13, 2012 #2
    This is much easier if you don't plug numbers in first.

    The reason you're getting the impossible answer is because you're assuming the peak of the curve is at the wall. You were asked to find the highest point the player could hit. You have to write an equation for the height of the ball as a function of angle and initial velocity, then find where it is highest.
  4. Oct 13, 2012 #3
    You need to split into two equations - horizontal and vertical. This way, you are able to solve this problem easily [takes some steps to solve the problem]

    First thing first.

    You need to determine the vertical component and the horizontal component of the path the ball travels, using the first formula you just indicated.

    Also, ask yourself:

    Which component of the path does gravity take place?
    Is there acceleration in horizontal, vertical or both?
  5. Oct 13, 2012 #4
    Gravity is in the vertical component of velocity, and acceleration is ZERO in the horizontal motion.

    Actually I was absent when my teacher solved this. I asked my friend and he did it using v^2=u^2+2as like this:

    ## \therefore \alpha=48.84 ## which is the correct answer.
    Why did they take s=6.5 m?
  6. Oct 13, 2012 #5
    Good. You said that there is no acceleration in the horizontal and gravity for the vertical component.

    Then, set up the expression, using this form...

    s = s_0 + vt + 0.5at^2 [where s_0 is your initial height]

    You get two equations - horizontal and vertical.
  7. Oct 13, 2012 #6
    Oh yeah I got it. frogjg was right, my assumption was wrong. I used s=ut+0.5at^2 and I got it!! Thank you!!
  8. Oct 13, 2012 #7
    v_f² = v_i² + 2as makes no use of what you are trying to show here.
  9. Oct 14, 2012 #8
    But my teacher used that and got the correct angle xD
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