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Question 1038562: When we kick a ball ,is it sure that ball always makes parabola if there is no air resistance?If yes,how can we find focus?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
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You need the magnitude and direction of the initial velocity vector. In the figure above,  represents the direction and  represents the magnitude of the initial velocity vector  . I'm going to assume that you are using the mks system.
The first thing that you need is to resolve the initial velocity vector into its horizontal and vertical components to get the magnitudes of the initial velocity in the horizontal direction and in the vertical direction. Use the following formulae where  is the magnitude of the vertical component and  is the magnitude of the horizontal component.
Since you are ignoring air resistance, we can presume that once the impulse of the kick has been applied, the only external force that acts upon the ball is gravity. Gravity only acts in the vertical, so it is the vertical velocity component that becomes a part of the altitude function. Let  represent the time in seconds since the ball was kicked.
\ =\ -4.9t^2\ +\ \left\|\vec{v_y_o}\right\|t\ +\ a_o)
If we also assume that the ball was resting on the ground when it was kicked, the initial altitude component,  , is zero and can be ignored. Thus:
\ =\ -4.9t^2\ +\ \left\|\vec{v_y_o}\right\|t)
The graph of the altitude function is a parabola, but it is not the parabola that you seek. The above is altitude as a function of time, whereas the function that you seek is altitude as a function of horizontal distance. However, in order to determine the horizontal distances, we need to know the time it takes the ball to reach its maximum height. This time is the value of at the vertex of the ) parabola. Recall that the  value of the vertex point of a parabola with equation  is given by  . Hence:
The maximum height of the ball is then the value of the altitude function at time  , in other words:
\ =\ -4.9t_{max}^2\ +\ \left\|\vec{v_y_o}\right\|t_{max})
The horizontal distance travelled by the ball at the time that the ball reaches the vertex of the parabola is simply [tex \Large t_{max}] multiplied by the horizontal component of the initial velocity, to wit:
\ =\ \left\|\vec{v_x_o}\right\|t_{max})
Now we have the coordinates of the vertex of our desired parabola, namely ,a(t_{max})\right)) . Since we assume the ball was on the ground at a horizontal distance of zero when the ball was kicked, we also know that the point ) is on our parabola.
Let us recall the equation of a parabola with a vertical axis:
^2\ +\ k)
where ) is the vertex of the parabola and  , the distance from the vertex to the focus is related to  by 
Let's see what all this means given that the point is on our parabola.
^2\ +\ k)
Therefore, the focus is at:
,a(t_{max})\ -\ \frac{\[h(t_{max})\]^2}{4\cdot a(t_{max})}\right))
Now all you need to do is the arithmetic. Good luck.
John
My calculator said it, I believe it, that settles it
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