# Two dimensional projectile motion

1. Apr 23, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
A regular hexagon of side a=10√3 m is kept at horizontal surface as shown in the figure. A particle is projected with velocity V m/s at an angle θ from point k such that it will just touch the all four corners B,C,D,E. If O is the centre of hexagon and x-axis is parallel to horizontal and y-axis is perpendicular to AF and CD sides of hexagon then answer the following

A) Velocity of projectile at maximum height

3. The attempt at a solution
Let k be the origin. R be the range of projectile
Coordinates of B = (R/2 - a, a√3/2)
Coordinates of C = (R/2 - a/2, a√3)
Equation of trajectory : $y=x \tan \theta \left(1-\dfrac{x}{R} \right)$

Satisfying the above coordinates does not give me the right answer.

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2. Apr 23, 2014

### Saitama

I don't think your coordinates are correct. Can you show how did you find them? Its difficult to follow your working.

3. Apr 23, 2014

### utkarshakash

Let the mid-point of AF be T. KT=R/2. Drop a perpendicular from B to the line KAFL. Call it BW.
AT=a/2. AW=acos60° = a/2. Thus, WT = a. x coordinate of B = KT-WT = R/2 - a.
For y-coordinate, BW = asin60 = a√3/2. Similar approach can be followed to find to coordinates of C.

4. Apr 23, 2014

### ehild

R is not known. Determine the coordinates from the given side of the hexagon, in he suggested coordinate system (origin is at the centre of the hexagon).

ehild

5. Apr 23, 2014

### Saitama

Looks ok to me.

Plugging in the coordinates, you get two equations, divide them to eliminate tan. Solve for R.

6. Apr 23, 2014

### utkarshakash

But then the equation of trajectory would need a transformation and it would be difficult to do that.

7. Apr 23, 2014

### ehild

The trajectory is a parabola, and you do not need the range. You need theta and V.

And you need the coordinates anyway.

ehild