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Release point

  1. Mar 1, 2006 #1
    A student throws a ball with a speed of 15.0 m.s-1 at an angle of 40.0° above the horizontal directly toward a wall as shown below. The wall is 10.0 m from the release point of the ball.

    Now I did figure this out. The vertical component of velocity (initial) is v= v(inital) sin angle, and in the vertical direction the ball has the constant acceleration due to gravity. So v = v(initial) sin angle - gt should be true for any given time.

    I can't figure out how far above the release point does the ball hit the wall?
     
  2. jcsd
  3. Mar 1, 2006 #2
    Let's start by listing all the possible equations you can use. There are 5.

    There's no acceleration in the x direction so:
    [tex]x=x_0+v_{0x}t[/tex]

    In the y direction:
    [tex]y=y_0+v_{0y}t+(1/2)a_yt^2[/tex]
    [tex]y=y_0+(1/2)(v_{0y}+v_y)t[tex]
    [tex]v_y=v_{0y}+a_yt[/tex]
    [tex]v_y^2=v_{0y}^2+2a_y(y-y_0)[/tex]

    So sketch your diagram. I'd recommend setting the origin where the ball is released and +x to the right, and +y upward.

    Which equation do you need to use to find how high the ball strikes the wall? What do you need to solve it?

    -Dan
     
  4. Mar 1, 2006 #3
    If you found how long is the ball in the air before it hits the wall, it can be used to answer your question. As you said the vertical position of the ball at anytime will be y = (v0 sin)t – ½ gt2. So if you find the time it will be a simple matter of substitution into the equation.

    Hope this will help.
     
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