Two dimensional projectile motion

[utkarshakash]

Homework Statement

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

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 : [itex]y=x \tan \theta \left(1-\dfrac{x}{R} \right) [/itex]

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

 

[Saitama]

Homework Statement

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

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 : [itex]y=x \tan \theta \left(1-\dfrac{x}{R} \right) [/itex]

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

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

 

[utkarshakash]

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

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.

 

[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

 

[Saitama]

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.

Looks ok to me.

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

 

[utkarshakash]

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

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

 

[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